IN   MEMORIAM 
FLOR1AN  CAJORI 


*"'• 


WORKS  OF  PROF.  G.  B.  HALSTED 


PUBLISHED    BY 


JOHN   WILEY   &   SONS. 


Elements  of  Geometry. 

8vo,  cloth,  $1.75. 

Synthetic  Geometry  . 

For  Schools  and  High  Schools.    8vo,  cloth,  $1.50. 

Raticnal  Geometry. 

A  Text-book  for  the  Science  of  Space.  Based  on 
Hilbert's  Foundations,  nmo,  viii  -f-  285  pages,  247 
figures.     Cloth,  $1.75. 


RATIONAL     GEOMETRY 


A  TEXT-BOOK  FOR 


THE   SCIENCE   OF   SPACE 


BASED  ON   HILBERT'S   FOUNDATIONS 


BY 

GEORGE  BRUCE  HALSTED 

A.B.  and  A.M.,  Princeton;  Ph.D.,  Johns  Hopkins 


FIRST  EDITION' 
FIRST    THOUSAND 


NEW  YORK 
'jOHN   WILEY   &   SONS 
London  :  CHAPMAN  &  HALL,  Limited 
1904 


&Ar46'3 


Copyright,  1904, 

BY 

GEORGE  BRUCE  HALSTED. 


CAJORl 


ROBERT  DRUMMOND,    PRINTER,    NHW  YORK. 


fcUf 


PREFACE. 


Writing  to  Professor  Hilbert  my  desire  to  base 
a  text-book  on  his  foundations,  he  answered :  M  Ueber 
Ihre  Idee  aus  meinen  Grundlagen  eine  Schul-Geo- 
metrie  zu  machen,  bin  ich  sehr  erfreut.  Ich  glaube 
auch,  dass  dieselben  sich  sehr  gut  dazu  eignen  wer- 
den." 

Geometry  at  last  made  rigorous  is  also  thereby 
made  more  simple. 

George  Bruce  Halsted. 

Kenyon  College, 
Gambier,  Ohio. 

iii 


9181iM 


CONTENTS. 


CHAPTER  I.  FAGE 

Association I 

CHAPTER  II. 
Betweenness 5 

CHAPTER  III. 
Congruence 15 

CHAPTER  IV. 
Parallels 35 


f  CHAPTER  V. 

'  The  Circle 49 


} 


CHAPTER  VI. 
Problems  of  Construction 62 

CHAPTER  VII. 
Sides,  Angles,  and  lArcs  A 73 

CHAPTER  VIII. 
A  Sect  Calculus 87 

CHAPTER  IX. 
Proportion  and  the  Theorems  of  Similitude 98 


VI  CONTENTS. 

CHAPTER  X. 

PAGB 

Equivalence 109 

CHAPTER  XL 
Geometry  of  Planes 139 

CHAPTER  XII. 
Polyhedrons  and  Volumes 163 

CHAPTER  XIII. 
Tridimensional  Spherics 194 

CHAPTER  XIV. 
Cone  and  Cylinder 205 

CHAPTER  XV. 
Pure  Spherics 212 

CHAPTER  XVI. 
Angloids  or  Polyhedral  Angles 248 

APPENDIX  I. 
Proofs  of  Betweenness  Theorems 253 

APPENDIX  II. 
The  Compasses 259 

APPENDIX  III. 
The  Solution  of  Problems 262 


TABLE  OF  SYMBOLS. 


We  denote 
triangle  by  A ;   the  vertices  by  A,  B,  C; 
the  angles  at  A,  B,  C  by  a,  {3,  y; 
the  opposite  sides  by  a,  b,  c; 
the  altitudes  from  A,  B,  C  by  ha,  hb,  hc; 
the  bisectors  of  a,  /?,  y  by  ta,  tb,  U\ 
the  medians  to  a,  b,  c  by  ma,  nib,  mc; 
the  feet  of  ha,  hb,  he  by  D,  E,  F\ 
the  centroid  by  G\ 
the  orthocenter  by  H ; 
the  in-center  by  / ;  the  in-radius  by  r ; 
the  ex-centers  beyond  a,  b,  c  by  llt  It,  I3;    their  ex-radii 

by  r„  r„  rs; 
the  circumcenter  by  O;   the  circumradius  by  R; 
angle  by  ^;   angles  by  ^s; 

angle  made  by  the  rays  B A  and  BC  by  ^,4i?C; 
angle  made  by  the  rays  a  and  b  both  from  the  point  O  by 

^  (a,  b)  or  2£a&; 
bisector  by  bi';    circle  by  O;    circles  by  Os; 
circle  with  center  C  and  radius  r  by  ©C(r); 
congruent  by  =  ; 

equal  or  equivalent  by  completion  by    = ; 
for  example  [exempli  gratia]  by  e.v.  • 
greater  than  by  >  ; 
less  than  by   <  ;   minus  by   —  ; 
parallel  by  ||;   parallels  by  ||s; 

vil 


viii  TABLE  OF  SYMBOLS. 

parallelogram  by  \\g'm;  perimeter  [sum  of  sides]  by  p; 

perpendicular  by  j_ ;    perpendiculars  by  Is; 

plus  by  +  ; 

quadrilateral  by  quad'; 

right  by  r't; 

spherical  angle  by  ^  ; 

spherical  triangle  by  'a  ; 

similar  by  ~ ; 

symmetrical  by  •!• ; 

therefore  by  ,\ 


RATIONAL  GEOMETRY. 


CHAPTER  I. 

ASSOCIATION. 

The  Geometric  Elements. 

i.  Geometry  is  the  science  created  to  give  under 
standing  and  mastery  of  the  external  relations  of 
things;  to  make  easy  the  explanation  and  descrip- 
tion of  such  relations  and  the  transmission  of  this 
mastery. 

2.  Convention.  We  think  three  different  sorts  of 
things.  The  things  of  the  first  kind  we  call  points, 
and  designate  them  by  A,  B,  C,  .  .  .  ;  the  things  of 
the  second  system  we  call  straights,  and  designate 
them  by  a,  b,  c,  ...  ;  the  things  of  the  third  set 
we  call  planes,  and  designate  them  by  a,  /?,  y,  .  .  .  . 

3.  We  think  the  points,  straights,  and  planes  in 
certain  mutual  relations,  and  we  designate  these 
relations  by  words  such  as  "lie,"  "between,"  " par- 
allel," "  congruent." 

The  exact  and  complete  description  of  these  rela- 


2  ^  RATIONAL  GEOMETRY. 

tiqnsr  is  accomplished  by  means  of  the  assumptions 
|©f  •gqcimetry: 

4.  The  assumptions  of  geometry  separate  into  five 
groups.  Each  of  these  groups  expresses  certain  con- 
nected fundamental  postulates  of  our  intuition. 

I.    The    first    group   of  assumptions:    assumptions    of 
association. 

5.  The  assumptions  of  this  group  set  up  an  asso- 
ciation between  the  concepts  above  mentioned, 
points,  straights,  and  planes.     They  are  as  follows: 

I  1.  Two  distinct  points,  A,  B,  always  determine 
a  straight,  a. 

Of  such  points  besides  " determine"  we  also  em- 
ploy other  turns  of  phrase;  for  example,  A  "lies 
on"  a,  A  "is  a  point  of"  a,  a  "goes  through"  A 
"and  through"  B,  a  "joins"  A  "and"  or  "with" 
B,  etc. 

When  we  say  two  things  determine  some  other 
thing,  we  simply  mean  that  if  the  two  be  given, 
then  this  third  is  explicitly  and  uniquely  given. 

If  A  lies  on  a  and  besides  on  another  straight  b 
we  use  also  the  expression :  ' '  the  straights  "  a  "  and ' ' 
b  "have  the  point  A  in  common." 

I  2.  Any  two  distinct  points  of  a  straight  determine 
this  straight;  and  on  every  straight  there  are  at  least 
two  points. 

That  is,  if  AB  determine  a  and  AC  determine 
a,  and  B  is  not  C,  then  also  B  and  C  determine  a. 

I  3.  Three  points,  A,  B,  C,  not  costraight,  always 
determine  a  plane  a. 


ASSOCIATION.  3 

We  use  also  the  expressions: 

A,  B,  C  "lie  in"  a,  A,  B,  C,  "are  points  of"  a, 
etc. 

I  4.  Any  three  non-costraight  points  A,  B,  C  of  a 
plane  a  determine  this  plane  a. 

I  5.  //  two  points  A,  B  of  a  straight  a  lie  in  a 
plane  a,  then  every  point  of  a  lies  in  a. 

In  this  case  we  say:  The  straight  a  lies  in  a. 

16.  If  two  planes  a,  /?  have  a  point  A  in  common, 
then  they  have  besides  at  least  another  point  B  in 
common. 

I  7.  In  every  plane  there  are  at  least  three  non- 
costraight  points.  There  are  at  least  four  non-co- 
straight non-coplanar  points. 

6.  Theorem.  Two  distinct  straights  cannot  have 
two  points  in  common. 

Proof.  The  two  points  being  on  the  first  straight 
determine  (by  I  2)  that  particular  straight.  If  by 
hypothesis  they  are  also  on  a  second  straight, 
therefore  (by  I  2)  they  determine  this  second 
straight.  Therefore  the  first  straight  is  identical 
with  the  second. 

7.  Theorem.  Two  straights  have  one  or  no  point 
in  common. 

Proof.     By  6  they  cannot  have  two. 

8.  Theorem.  Two  planes  have  no  point  or  a 
straight  in  common. 

Proof.  If  they  have  one  point  in  common,  then 
(by  I  6)  they  have  a  second  point  in  common,  and 
therefore  (by  I  5)  each  has  in  it  the  straight  which 
(by  I  1)  is  determined  by  these  two  points. 

9.  Corollary  to  8.     A  point  common  to  two  planes 


4  RATIONAL   GEOMETRY. 

lies  in  a  straight  common  to  the  two,  which  may  be 
called  their  straight  of  intersection  or  their  meet. 

10.  Theorem.  A  plane  and  a  straight  not  lying  in 
it  have  no  point  or  one  point  in  common. 

Proof.  If  they  had  two  points  in  common  the 
straight  would  be  (by  I  5)  situated  completely  in 
the  plane. 

11.  Theorem.  Through  a  straight  and  a  point  not 
on  it  there  is  always  one  and  only  one  plane. 

Proof.  On  the  straight  there  are  (by  I  2)  two 
points.  These  two  with  the  point  not  on  the 
straight  determine  (by  I  3)  a  plane,  in  which  (by 
I  5)  they  and  the  given  straight  lie.  Any  plane 
on  this  point  and  straight  would  be  on  the  three 
points  already  used,  hence  (by  I  4)  identical  with 
the  plane  determined. 

12.  Theorem.  Through  two  different  straights  with 
a  common  point  there  is  always  one  and  only  one 
plane. 

Proof.  Each  straight  has  on  it  (by  I  2)  one 
point  besides  the  common  point,  and  (by  6)  these 
two  points  are  not  the  same  point,  and  (by  I  2) 
the  three  points  are  not  costraight. 

These  three  points  determine  (by  I  3)  a  plane  in 
which  (by  I  5)  each  of  the  two  straights  lies.  Any 
plane  on  these  straights  would  be  on  the  three 
points  already  used,  hence  (by  I  4)  identical  with 
the  plane  determined. 


CHAPTER  II. 

BETWEENNESS. 

II.  The  second  group  of  assumptions:   assumptions  of 
betweenness. 

13.  The  assumptions  of  this  group  make  precise 
the  idea  "between,"  and  make  possible  on  the  basis 
of  this  idea  the  arrangement  of  points. 

14.  Convention.  The  points  of  a  straight  stand 
in  certain  relations  to  one  another,  to  describe 
which  especially  the  word  "between"  serves  us. 

II  1.  If  A,  B,  C  are  points  of  a  straight,  and  B  lies 
between  A  and  C,  then  B  also  lies  between  C  and  A, 
and  is  neither  C  nor  A. 


4 § ?■ 


Fig. 


II  2.  If  A  and  C  are  two  points  of  a  straight,  then 
there  is  always  at  least  one  point  B,  which  lies  between 


Fig. 


A  and  C,  and  at  least  one  point  D,  such  that  C  lies 
between  A  and  D. 

5 


6  RATIONAL   GEOMETRY 

II  3.  Of  any  three  points  of  a  straight  there  is 
always  one  and  only  one  which  lies  between  the  other 
two. 

15.  Definition.  Two  points  A  and  B,  upon  a 
straight  a,  we  call  a  segment  or  sect,  and  designate 
it  with  AB  or  BA.  The  points  between  A  and  B 
are  said  to  be  points  of  the  sect  AB  or  also  situated 
within  the  sect  AB.  All  remaining  points  of  the 
straight  a  are  said  to  be  situated  without  the  sect 
AB.  The  points  A,  B  are  called  end-points  of  the 
sect  AB. 

II  4.  (Pasch's  assumption.)  Let  A,  B,  C  be  three 
points  not  costraight  and  a  a  straight  in  the  plane 
ABC  going  through  none  of  the  points  A,  B,  C;   if 


Fig.  3. 

then  the  straight  a  goes  through  a  point  within  the 
sect  AB,  it  must  always  go  either  through  a  point  of 
the  sect  BC  or  through  a  point  of  the  sect  AC. 

Deductions   from  the    assumptions  of   association   and 
betweenness. 

16.  Theorem.     Between  any  two  points  of  a  straight 
there  are  always  indefinitely  many  points. 

[Here  taken  for  granted,  and  its  proof  removed  to 
Appendix  I.] 

17.  Theorem.     If  any  finite  number  of  points  of 


BETIVEENNESS.  7 

a  straight  are  given,  then  they  can  always  be  ar- 
ranged in  a  succession  A,  B,  C,  D,  E,  .  .  .  ,  K,  such 
that  B  lies  between  A  on  the  one  hand  and  C,  D, 
E,  .  .  .  ,  AT  on  the  other,  further  C  between  A,  B 
on  the  one  hand  and  D,  Et  .  .  .  ,  K  on  the  other, 
then  D  between  A,  B,  C  on  the  one  hand  and 
E}  .  .  .  ,  K  on  the  other,  and  so  on. 

Besides  this  distribution  there  is  only  one  other, 
the  reversed  arrangement,  which  is  of  the  same 
character. 

[This  theorem  is  here  taken  for  granted,  and  its 
proof  removed  to  Appendix  I.] 

21.  Theorem.  If  A,  B,  C  be  not  costraight,  any 
straight  in  the  plane  ABC  which  has  a  point  within 
the  sect  AB  and  a  point  within  AC  cannot  have  a 
point  within  BC. 

Proof.  Suppose  F,  G,  H  three  such  costraight 
points. 

One,  say  G,  on  AB,  must  (by  II  3)  lie  between  the 
others.  Then  the  straight  AB  must  (by  II  4)  have 
a  point  within  the  sect  EC  or  the  sect  CH,  which 
(by  7  and  II  3)  is  impossible. 


Fig.  4. 

22.  Theorem.  Every  straight  a,  which  lies  in  a 
plane  a,  separates  the  other  points  of  this  plane  a 
into  two  regions,   of  the  following  character:    every 


RATIONAL   GEOMETRY. 


point  A  of  the  one  region  determines  with  every  point 
B  of  the  other  region  a  sect  AB,  within  which  lies  a 
point  of  the  straight  a;    on  the  contrary,   any  two 


Fig.  5. 

points  A,  Af  of  one  and  the  same  region  always  deter  - 
mine  a  sect  A  A'  which  contains  no  point  of  a. 

Proof.  Let  A  be  a  point  of  a  which  does  not  lie 
on  a.  Then  reckon  to  one  region  all  points  P  of 
the   property,   that   between  A   and  P,    therefore 


Fig.  6. 

within  AP,  lies  no  point  of  a;   to  the  other  region 
all  points  Q  such  that  within  AQ  lies  a  point  of  a. 
Now  is  to  be  shown : 

(1)  On  PP'  lies  no  point  of  a. 

(2)  On  QQ'  lies  no  point  of  a. 

(3)  On  PQ  lies  always  a  point  of  a. 


BETU/EENNESS.  9 

(i)  From  hypothesis  neither  within  AP  nor  AP' 
lies  a  point  of  a.  This  would  contradict  II  4,  if 
within  PP'  were  a  point  of  a. 

(2)  By  hypothesis  there  lies  within  AQ  a  point 
of  a,  likewise  within  AG' \  therefore  (by  21)  none 
within  QQ'. 

(3)  By  hypothesis  AP  contains  no  point  of  a; 
AQ  on  the  other  hand  contains  one  such.  There- 
fore (by  II  4)  a  meets  PQ. 

23.  Convention.  If  A,  A'}0}  B  are  four  costraight 
points  such  that  0  is  between  A  and  B  but  not 
between  A  and  ^4';  then  we  say:  the  points  A,  Af 


Fig.  7. 

lie  w  //t£  straight  a  on  one  and  the  same  side  of  the 
point  0,  and  the  points  A,  B  lie  in  the  straight  a  on 
different  sides  of  the  point  0. 

24.  Definition.  The  assemblage,  aggregate,  or  to- 
tality of  all  points  of  the  straight  a  situated  on  one 
and  the  same  side  of  O  is  called  a  ray  starting  from  0. 

Consequently  every  point  of  a  straight  is  the  ori- 
gin of  two  rays. 

25.  Convention.  Using  the  notation  of  22,  we 
say:  the  points  P,  P'  lie  in  the  plane  a  on  one  and 
the  same  side  of  the  straight  a  and  the  points  P,  Q  lie 
in  the  plane  a  on  different  sides  of  the  straight  a. 

26.  Theorem.  Every  two  intersecting  straights  a, 
b  separate  the  points  of  their  plane  a  not  on  either 
into  four  regions  such  that  if  the  end-points  of  a  sect 
are  both  in  one  of  these  regions,  the  sect  contains  no 
point  of  either  straight. 


io  RATIONAL   GEOMETRY. 

Proof.  Let  0  be  their  common  point  and  A 
another  point  on  b,  and  B  another  point  on  a. 
Then  two  points  both  on  the  A  side  of  a  and  the 
B  side  of  b  make  a  sect  which  (by  22)  can  contain 
no  point  either  of  a  or  of  b.     So  also  if  both  were 


Fig.  8. 

on  the  A  side  of  a  and  the  non-5  side  of  b ;  or  both 
on  the  non-A  side  of  a  and  the  B  side  of  b ;  or  both 
on  the  non-.4  side  of  a  and  the  non-B  side  of  b. 

27.  Definition.  A  system  of  sects  AB,  BC, 
CD,  .  .  .  ,  KL  is  called  a  sect-train,  which  joins  the 
points  A  and  L  with  one  another.  This  sect-train 
will  also  be  designated  for  brevity  by  A  BCD  .  .  .  KL. 

The  points  within  the  sects  AB,  BC,  CD,  .  .  .  ,  KL, 
together  with  the  points  A,  B,  C,  D,  .  .  .  ,  K,  L 
are  all  together  called  the  points  of  the  sect-train. 

In  particular  if  the  point  L  is  identical  with  the 
point  A,  then  the  sect-train  is  called  a  polygon  and 
is  designated  as  polygon  A  BCD  .  .  .  K. 

The  sects  AB,  BC,  CD,  .  .  .  ,  KA  are  called  the 
sides  of  the  polygon.  The  points  A,  B,  C,  D,  .  .  . , 
K  are  called  the  vertices  of  the  polygon. 

A  sect  not  a  side  but  whose  end-points  are  ver- 
tices is  called  a  diagonal  of  the  polygon. 


BETIVEENNESS.  n 

Polygons  with  3,  4,  5,  .  .  .  ,  n  vertices  are  called 
respectively  triangles,  quadrilaterals,  pentagons,  .  .  .  , 
n-gons. 

28.  If  the  vertices  of  a  polygon  are  all  distinct 
from  one  another  and  no  vertex  of  the  polygon 
falls  within  a  side  and  finally  no  two  sides  of  the 
polygon  have  a  point  within  in  common,  then  the 
polygon  is  called  simple. 

By  quadrilateral  is  meant  simple  quadrilateral. 

A  plane  polygon  is  one  all  of  whose  sides  are  co- 
planar. 

A  convex  polygon  is  one  no  points  of  which  a  e 
on  different  sides  of  the  straight  of  any  of  its  sides 

29.  Theorem.  Every  simple  polygon,  whose  ver- 
tices all  lie  in  a  plane  a,  separates  the  points  of  this 
plane  a,  which  do  not  pertain  to  the  sect-train  of 
the  polygon,  into  two  regions,  an  inner  and  an  outer, 
of  the  following  character:  if  A  is  a  point  of  the 
inner  {interior  point)  and  B  a  point  of  the  outer  < 
{exterior  point),  then  every  sect-train  which  joins 

A  with  B  has  at  least  one  point  in  common  with  the  ^3 
polygon;  on  the  contrary  if  A,  A'  are  two  points  bf 
of  the  inner  and  B,  B'  two  points  of  the  outer,  then  ^ 
there  are  always  sect-trains,  which  join  A  with  A'^ 
and  B  with  B'  and  have  no  point  in  common  with 
the  polygon. 

There  are  straights  in  a  which  lie  wholly  outside 
the  polygon;  on  the  contrary  no  such  straights 
which  lie  wholly  within  the  polygon. 

Proof.  Any  simple  polygon  by  joining  its  ver- 
tices gives  a  number  of  triangles.  For  a  triangle 
ABC  there  is  (by  26)  a  region  with  points  on  the 


12  RATIONAL   GEOMETRY. 

A  side  of  BC,  the  B  side  of  CA,  and  the  C  side  of 
A B,  i.e.,  an  inner  region.  Moreover,  the  straight 
determined  by  a  point  on  b  and  a  point  on  c  both  in 
non-A  lies  wholly  without  the  region  ABC,  since 
it  cannot  again  meet  b  or  c  and  so  cannot  (by  II  4) 


Fig.  9. 

have  a  point  in  common  with  BC.  Moreover,  if 
any  straight  has  a  point  within  ABC,  it  has  a. point 
on  a  side.  For  the  straight  determined  by  the 
point  within  and  any  point  on  a  side  has  (by  II  4) 
a  point  on  another  side,  thus  making  another  tri- 
angle, in  common  with  one  side  of  which  the  given 
straight  has  a  point,  and  therefore  (by  II  4)  with 
another  side,  that  is  with  a  side  of  the  original  tri- 
angle. 

30.  Corollary  to  29.  A  straight  through  a  ver- 
tex and  a  point  within  a  triangle  has  a  point  within 
the  'opposite'  side. 


BETIVEENNESS.  13 

31.  Theorem.  Every  plane  a  separates  all  points 
not  on  it  into  two  regions  of  the  following  character : 
every  point  A  of  the  one  region  determines  with 
every  point  B  of  the  other  region  a  sect  A  B,  within 
which  lies  a  point  of  a\  on  the  contrary  any  two 
points  A  and  A'  of  one  and  the  same  region  always 
determine  a  sect  A  A',  which  contains  no  point  of  a. 

Proof.  Let  A  be  a  point  which  does  not  lie  on 
a.  Then  reckon  to  the  one  region  all  points  P  of 
the  property,  that  between  A  and  P,  therefore  with- 
in AP,  lies  no  point  of  a;  to  the  other  region  all 
points  Q  such  that  within  AQ  lies  a  point  of  a. 

Now  is  to  be  shown: 

(1)  On  PP'  lies  no  point  of  a. 

(2)  On  QQ'  lies  no  point  of  a. 

(3)  On  PQ  lies  always  a  point  of  a. 

(1)  From  hypothesis  neither  within  AP  nor  APf 
lies  a  point  of  a.  Suppose  now  a  point  of  a  lay  on 
'PP' .  Then  the  plane  a  and  the  plane  APP'  would 
have  in  common  this  point  and  consequently  (by 
9)  a  straight  a.  This  straight  goes  through  none 
of  the  points  A,  P,  P';  it  cuts  PP' ;  it  must  there- 
fore (by  II  4)  cut  either  AP  or  AP',  which  is  con- 
trary to  hypothesis. 

(2)  By  hypothesis  there  lies  within  AQ  &  point 
of  a,  likewise  within  AQ'.  The  intersection  straight 
of  the  planes  a  and  AQQ'  therefore  meets  two  sides 
of  the  triangle  A  QQ' ;  consequently  (by  21)  it  can- 
not also  meet  the  other  side  QQ'. 

(3)  AP  contains  by  hypothesis  no  point  of  a; 
AQ  on  the  other  hand  contains  one  such.  The  inter- 
section straight  of  the  planes  a  and  APQ  therefore 


14  RATIONAL   GEOMETRY. 

meets  the  side  AQ  and  does  not  meet  the  side  AP 
in  triangle  APQ.  Therefore  (by  II  4)  it  meets  the 
side  PQ. 

32.  Convention.  Using  the  notation  of  31,  we 
say:  the  points  A,  A'  lie  on  one  and  the  same  side 
of  the  plane  a,  and  the  points  A,  B  lie  on  different 
sides  of  the  plane  a. 

Ex.  i.  A  straight  cannot  traverse  more  than  4  of  the 
7  regions  of  the  plane  determined  by  the  straights  of  the 
sides  of  a  triangle. 

Ex.  2.  Four  coplanar  straights  crossing  two  and  two 
determine  6  points.  Choosing  4  as  vertices  we  can  get 
two  convex  quadrilaterals,  one  of  which  has  its  sides  on 
the  straights. 

Ex.  3.  Each  vertex  of  an  n-gon  determines  with  th : 
others  (n  —  1)  straights.  So  together  they  determine 
n(n  — 1)/2. 

Ex.  4.  How  many  diagonals  in  a  polygon  of  n  sides. 

Ex.  5.  What  polygon  has  as  many  diagonals  as  sides? 


-  r 


CHAPTER  III. 

CONGRUENCE. 

III.    The  third   group  of  assumptions:    assumptions  of 
congruence. 

33.  The  assumptions  of  this  group  make  precise 
the  idea  of  congruence. 

34.  Convention.  Sects  stand  in  certain  rela- 
tions to  one  another,  for  whose  description  the 
word  congruent  especially  serves  us. 

Ill  1.  If  A,  B  are  two  points  on  a  straight  a,  and 
A'  a  point  on  the  same  or  another  straight  a' ,  then 
we  can  -find  on  the  straight  a!  on  a  given  ray  from 
A'  always  one  and  only  one  point  B'  such  that  the 
sect  AB  is  congruent  to  the  sect  A'B'. 

We  write  this  in  symbols  AB  =  A'B'. 

Every  sect  is  congruent  to  itself,  i.e. ,  always  A  B  =  AB. 
The  sect  AB  is  always  congruent  to  the  sect  BA,  i.e., 
AB  =  BA. 

We  also  say  more  briefly,  that  every  sect  can  be 
taken  on  a  given  side  of  a  given  point  on  a  given 
straight  in  one  and  only  one  way. 

Ill  2.  If  a  sect  AB  is  congruent  as  well  to  the  sect 
A'B'  as  also  to  the  sect  A"B",  then  is  also  A'B'  con- 
is 


1 6  RATIONAL   GEOMETRY. 

gruent   to    the   sect   A"B",    i.e.,    if   AB  =  A'B'    and 
AB^A"B",  then  is  also  A,B,^-A,,B". 

Ill  3.  On  the  straight  a  let  AB  and  BC  be  two  sects 
without  common  points,  and  furthermore  A'B'  and 
B'C  two  sects  on  the  same  or  another  straight,  like- 
wise without  common  points;  if  then  AB  =  A'B'  and 
BC  =  B'C,  so  always  also  AC  =  A'C 


Fig.  10. 

35.  Definition.  Let  a  be  any  plane  and  h,  k  any 
two  distinct  rays  in  a  going  out  from  a  point  0,  and 
pertaining  to  different  straights.  These  two  rays 
h,  k  we  call  an  angle,  and  des- 
ignate it  by^  (h,  k)  or  4  (k,  h)- 
The  rays  h  and  k,  together  with 
the  point  0,  separate  the  other 
points  of  the  plane  a  into  two 
regions  of  the  following  character : 
if  A  is  a  point  of  the  one  region 

and  B  of  the  other  region,  then 
Fig.  11.  ......... 

every  sect-tram  which  joins  A  with 

B,  goes  either  through  0  or  has  with  h  or  k  at  least 

one  point  in  common;  on  the  contrary  if  A,  A'  are 

points  of  the  same  region,  then  there  is  always  a 

sect-train  which  joins  A  with  A'  and  neither  goes 

through  0  nor  through  a  point  of  the  rays  h,  k. 

One  of  these  two  regions  is  distinguished  from 

the  other  because  each  sect  which  joins  any  two 

points  of  this  distinguished  region  always  lies  wholly 


CONGRUENCE.  1 7 

in  it ;  this  distinguished  region  is  called  the  interior 
of  the  angle  (h,  k)  in  contradistinction  from  the 
other  region,  which  is  called  the  exterior  of  the 
angle  (h,  k).  The  interior  of  ^  (h,  k)  is  wholly  on 
the  same  side  of  the  straight  h  as  is  the  ray  k,  and 
altogether  on  the  same  side  of  the  straight  k  as  is 
the  ray  h. 

The  rays  h,  k  are  called  sides  of  the  angle,  and 
the  point  0  is  called  the  vertex  of  the  angle. 

Ill  4.  Given  any  angle  (h,  k)  in  a  plane  a  and  a 
straight  a'  in  a  plane  a! ,  also  a  determined  side  of  a' 
on  a' .  Designate  by  h'  a  ray  of  the  straight  a'  start- 
ing from  the  point  0' ;  then  there  is  in  the  plane  af 
one  and  only  one  ray  k'  such  that  the  angle  (h,  k) 
is  congruent  to  the  angle  (h',  k')y  and  likewise  all  in- 
terior points  of  the  angle  {h\  k')  lie  on  the  given  side 
of  a'. 

k 


In  symbols: 

Every    angle   is   congruent    to    itself,    i.e.,    always 

The  angle  (h,  k)  is  always  congruent  to  the  angle 
(AU),  i.e.,    $<&*)■*(*.*). 

We  say  also  briefly,  that  in  a  given  plane  every 
angle  can  be  set  off  towards  a  given  side  against  a 


1 8  RATIONAL    GEOMETRY. 

given  ray,  but  in  a  uniquely  determined  way. 
There  is  one  and  only  one  such  angle  congruent  to 
a  given  angle.  We  say  an  angle  so  taken  is  uniquely 
determined. 

Ill  5.  //  an  angle  (h,  k)  is  congruent  as  well  to 
the  angle  (h\  k')  as  also  to  the  angle  (h"}  k"),  then  is 
also  the  angle  (h't  k')  congruent  to  the  angle  (h",  k") ; 
i.e.,  if  *  (h,  k)=if  (h't  V)  and  *  (h,  k)^4  (h",  k"), 
then  always  4  (h\  k')  =  %.  (h"t  k"). 

36.  Convention.  Let  ABC  be  any  assigned  tri- 
angle; we  designate  the  two  rays  going  out  from 
A  through  B  and  C  respectively  by  h  and  k.  Then 
the  angle  (h,  k)  is  called  the  angle  of  the  triangle 
ABC  included  by  the  sides  A B  and  AC  or  opposite 
the  side  BC.  It  contains  in  its  interior  all  the  inner 
points  of  the  triangle  ABC  and  is  designated  by 
$BAC  or  4  A. 

Ill  6.  If  for  two  triangles  ABC  and  A'B'O  we 
have  the  congruences 

AB^A'B',    AC^A'C,     4  BAC  m  4  B'A'C, 

then  always  are  fulfilled  the  congruences 

4  ABC  m  4  A'B'C     and     *  ACB  s  4  A'C'B'. 

Deductions  from  the  assumptions  of  congruence. 

37.  Convention.  Suppose  the  sect  AB  congruent 
to  the  sect  A'B':  Since,  by  assumption  III  t,  also 
the  sect  AB  is  congruent  to  AB,  so  follows  from 
III  2  that  A'B'  is  congruent  to  AB;  we  say:  the 
two  sects  AB  and  A'B'  are  congruent  to  one  another. 

38.  Convention.     Suppose  a£  (h,  k)  m  -4.  (h't  k'). 


CONGRUENCE.  1 9 

Since  (by  III  4)  4  (h,  k)=4  (h,  k),  therefore 
(by  III  5)  *  (h/  k')  =  7f  (h,  k).  We  say  then:  the 
two  angles  4  Qh  k)  and  ^  (#,  &')  are  congruent  to 
one  another. 

39.  Definition.  Two  angles  having  the  same  ver- 
tex and  one  side  in  common,  while  the  sides  not 
common  form  a  straight,  are  called  adjacent  angles. 

40.  Definition.  Two  angles  with  a  common  ver- 
tex and  whose  sides  form  two  straights  are  called 
vertical  angles. 

41.  Definition.  Any  angle  which  is  congruent  to 
one  of  its  adjacent  angles  is  called  a  right  angle. 

Two  straights  which  make  a  right  angle  are  said 
to  be  perpendicular  to  one  another. 

42.  Convention.  Two  triangles  ABC  and  A'B'C 
are  called  congruent  to  one  another,  if  all  the  con- 
gruences 

AB  =  A'B\     AC**A'Ct     BC  =  B'C, 
lA=lA't     IB^IB',     IC^tC 

are  fulfilled. 

43.  (First  congruence  theorem  for  triangles.) 
Triangles  are  congruent  if  they  have  two  sides  and 

the  included  angle  congruent. 


In  the  triangles  ABC  and  A'B'C  take  AB  =  A'B\ 
AC  =  A'C\  4A=^-A'. 


20  RATIONAL    GEOMETRY. 

To  prove  aABC  =  aA'B'C. 

Proof.  By  assumption  III  6  the  congruences 
ifB=  ~^B'  and  ^C=  ~4-C  are  fulfilled,  and  so  we 
have  only  to  show  that  the  sides  BC  and  B'C  are 
congruent  to  one  another. 

Suppose  now,  on  the  contrary,  that  BC  were  not 
congruent  to  B'C \  and  take  on  ray  B'C  (by  III  i) 
the  point  D't  such  that  BC  =  B'U.  Then  the  two 
triangles  ABC  and  A'B'T)'  will  have,  since  ^-B  = 
^-B',  two  sides  and  the  included  angle  respectively 
congruent;  by  assumption  III  6,  consequently,  are 
in  particular  the  two  angles  B AC  and  B'A'D*  con- 
gruent to  one  another.  By  assumption  III  5,  con- 
sequently, must  therefore  also  the  two  angles  B' A'C! 
and  B' A'T)'  be  congruent  to  one  another.  This  is 
impossible,  since,  by  assumption  III  4,  against  a 
given  ray  toward  a  given  side  in  a  given  plane 
there  is  only  one  angle  congruent  to  a  given  angle. 
So  the  theorem  is  completely  established. 

44.  (Second  congruence  theorem  for  triangles.) 

Two  triangles  are  congruent  if  a  side  and  the  two 
adjoining  angles  are  respectively  congruent. 


Fig.  14. 

In  the  triangles  ABC  and  A' B'C  take  AC  =  A'C'% 
^A^^A'%   i-C^i-C. 
To  prove  aABC^  aA'B'C  .     . 


CONGRUENCE.  21 

Proof.  Suppose  now,  on  the  contrary,  AB  is  not 
=  A'B',  and  take  on  ray  A'B'  the  point  D't  such 
that  AB^A'U.  By  III  6,  *ACB  =  ^A'CD',  but 
by  hypothesis  ^ACB^  ^A'CB'.  Therefore  (by 
III  5)  1A'C'B'=  I  A' CD'.  But  this  is  impossible, 
since  (by  III  4)  in  a  given  plane  against  a  given  ray 
toward  a  given  side  there  is  only  one  angle  con- 
gruent to  a  given  angle. 

Consequently  our  supposition,  AB  not  =  A 'B' ,  is 
false,  and  so  AB  =  A'B'. 

Now  follows  (by  43)  that   aABC=  aA'B'C. 

45.  Theorem.  //  two  angles  are  congruent,  so  are 
also  their  adjacent  angles. 

Take  ^ABC=  TfA'B'C. 

To  prove  ifCBD^  ^CB'D'. 

Proof.  Choose  the  points  A',  C,  D'  on  the  sides 
from  Bf  so  that  A'B'  =  AB,  CB'  =  CB,  DB  =  D'B'. 


Jn  the  two  triangles  ABC  and  A'B'C  then  the 
sides  AB  and  CB  are  congruent  respectively  to  the 
sides  A'B1  and  CB',  and  since  moreover  the  angles 
included  by  these  sides  are  congruent  by  hypothesis, 
so  follows  (by  43)  the  congruence  of  those  triangles, 
that  is,  we  have  the  congruences 

AC^A'C     and     i-BAC^iB' A'C . 


22  RATIONAL   GEOMETRY. 


Now  since  (by  III  3)  sect  AD  =  A'D\  so  follows 
(again  by  43)  the  congruence  of  the  triangles  CAD 
and  C'A'D',  that  is,  we  have  the  congruences  CD  = 
CD'  and  $  ADC  &  if  A' PC',  and  hence  follows, 
through  consideration  of  the  triangles  BCD  and 
B'C'D'  (by  III  6),  the  congruence  of  the  angles  CBD 
and  C'B'U. 

46.  Theorem.     Vertical  angles  are  congruent. 

Proof.  By  III  4,  ^ABC^^CBA.  Therefore, 
by  45,  their  adjacent  angles  are  congruent,  ^.CBD  = 
4ABF. 


47.  Theorem.  Through  a  point  A,  not  on  a  straight 
a,  there  is  one  and  only  one  perpendicular  to  a. 

Proof.  Take  any  two  points  P,  Q  on  a.  Take 
from  P  against  the  side  PQ  of  ifAPQ,  and  on  the 
non-A  side  of  a,  *fBPQ=  iAPQ.  Take  PB  =  PA, 
Since  A  and  B  lie  on  different  sides  of  a,  there  must 
be  a  point  O  of  sect  A B  on  a.  Then  AOB  is  perpen- 
dicular to  a. 

For  (by  43)  aBPO=  aAPO,  so  ^BOP^^fAOP. 
But  these  are  adjacent.  Therefore,  by  definition  41, 
A  OP  is  a  right  angle. 

Moreover  this  perpendicular  is  unique.  For  sup- 
pose any.  straight  AO'  perpendicular  to  a  at  0',  and 


CONGRUENCE. 


23 


take  on  this  straight  on  the  non-^4  side  of  a  the  sect 
0'B'  =  0'A.  Then  from  hypothesis  i{PO'B'  = 
IPO* A  and  so  (by  43)   aPO'B'=  aPO'A.     There- 


Fig.   17. 

fore  4  B'PO'  =^APO'  and  B'P ^AP.  Therefore 
(by  III  5  and  III  2),  ^B'PO'=  *  BPO  and  B'P  =  BP. 
Hence  the  points  B  and  B'  are  not  different.  There- 
fore no  second  perpendicular  from  A  to  a  can  exist. 

48.  Theorem.  Let  the  angle  (h,  k)  in  the  plane 
a  be  congruent  to  the  angle  (/*',  k')  in  the  plane  «', 
and  further  let  /  be  a  ray  of  the  plane  a,  which  goes 
out  from  the  vertex  of  the  angle  (h,  k)  and  lies  in 
the  interior  of  this  angle ;  then  there  is  always  a  ray 
/'  in  the  plane  a',  which  goes  out  from  the  vertex  of 
the  angle  {h\  k')  and  lies  in  the  interior  of  this  angle, 
such  that  i.  (h,  l)=^f  (h',  V)  and  *  (k,  l)=^f  (&',  /')• 

Proof.  Designate  the  vertex  of  if  (h,  k)  by  0,  and 
the  vertex  of  4  W*  k')  by  0',  and  then  determine  on 
the  sides  h,  k,  h\  kf,  the  points  A,  B,  A\  B',  so  that 
we  have  the  congruences 

OA^O'A'     and     0B  =  0'B'. 


24 


RATIONAL    GEOMETRY. 


Because  of  the   congruence   of   the   triangles  OAB 
and  O'A'B'  (by  43) 

AB  =  A'B',    ^OAB^^fO'A'B',    iOBA=  ^fO'B'A'. 
The  straight  AB  (by  30)  cuts  /,  say  in  C\  then 
we  determine  on  the  sect  &!B'  the  point  C,  such 
that  A'C'sAC,  then  is  O'C  the  ray  sought,  V . 


In  fact,  from  AC  =  A'C  and  AB  =  A'B'  we  may, 
by  means  of  III  3,  deduce  the  congruence  BC  =  B'C. 
Therefore  (by  III  6)  %.  AOC=  *  A' O'C'  and  ^BOC 
^^B'O'C. 

49.  Theorem.  Let  h,  k,  I  on  the  one  hand  and 
h',  k',  V  on  the  other  each  be  three  rays  going  out 
from  a  point  and  lying  in  a  plane ;  if  then  we  have 
the  congruences  ^  (h,  I)  =  ^  (h',  /')  and  ^  (k,  I)  m 
4  (k',  /')>  then  also  is  always 


CONGRUENCE. 


25 


Proof.  The  rays  are  supposed  such  that  either 
no  point  is  interior  to  ^-{h,  I)  and  ?£(k,  /),  or  to 
"4-{h\  I')  and  4  (&',  /'),  or  else  that  if  one  of  these 
angles  be  within  a  second,  then  the  angle  congruent 
to  the  first  is  within  the  fourth. 

I.  In  the  first  case,  if  /  be  supposed  within  4  (h,  k), 
take  against  h'  toward  k't  4  (/*',  k")  a  4  (h,  k).     By 


Fig.  19. 


48,  take  in  angle  (h't  k")  ray  I"  such  that  4  (h't  Z") 
=  l(h,l)  and  *(/",£")=*(/,&).  But  by  hy- 
pothesis i(h\  V)mt(h,f).  Therefore,  by  III  5, 
t(.  (/*',  V)  =  4  (fc',/"),  and  so,  by  III  4,  ray  I"  is  iden- 
tical with  ray  /'.  Then  4  (k",  /")  m  4  (£",  V)  = 
4  (&,  0  =  *  (&',  l')-  So  4  (k",  V)  s  4  (k'}  /'),  and,  by 
III  4,  ray  k"  is  identical  with  ray  k' '. 

But  4  (h\  k")  3  4  (h,  k) .  Therefore  4  (h,  k)  = 
4(h',k'). 

If,  however,  /  be  supposed  not  within  ^(/j,  &), 
then  it  will  lie  in  4  i)i" ',  k")  vertical  to  4  (h,  k).  For 
it  cannot  lie  in  4-{h,k")  adjacent  to  ^  (/*,  &),  since 
then  4  (I,  k)  would  contain  4  (h*  0>  contradicting  the 
hypothesis  in  this  case  of  no  point  interior  to  these 
two  given   angles.     For  like  reason  it  cannot  lie  in 


26 


RATIONAL   GEOMETRY. 


i£(h",  k)  adjacent  to  £(/*,  k),  since  then  4  (h,  1) 
would  contain  %.  (/,  fe).  Thus  the  ray  m  costraight 
with  I  is  within  i£(h,  k),  and  m'  costraight  with  V 
is  within  4  Q1',  k'). 


Fig.  20. 


Then  (by  45)  ^  (&,  w)  =  ^  (/*',  m')  and  ^  (&,  w)  = 
^-(k',w!)  [^'s  adjacent  to  congruent  %.  's  are  con- 
gruent], and  so  this  sub- case  is  reduced  to  the  pre- 
ceding. 

II.  The  remaining  case,  where  one  angle  £(h,  I) 
is  within  another,  s£  (k,  I),  follows  at  once  from  48. 

51.  Theorem.  All  right  angles  are  congruent. 
Let  angle  BAD  be  congruent  to  its  adjacent  angle 
CAD,  and  likewise  let  the  angle  B'A'Dr  be  con- 
gruent to  its  adjacent  angle  C'A'D'\  then  are 
■4.  BAD,  4  CAD,  4-B'A'D',  ^CA'D'  all  right 
angles. 

To  prove  4BAD=  fB'A'U. 

Proof.  Suppose,  contrary  to  our  proposition,  the 
right  angle  B'A'D'  were  not  congruent  to  the  right 
angle  BAD,  and  then  set  off  4-BrA'Dt  against  ray 


CONGRUENCE. 


27 


AB  so  that  the  resulting  side  AD"  falls  either  in  the 
interior  of  the  angle  BAD  or  of  the  angle  CAD ;  sup- 
pose we  have  the  first  of  these  cases. 

Because         $B' 'A'D' '=  4  BAD" \  therefore, 

by  45,  * C'A'D'  =  i CAD" ;        and  since  by 

hypothesis        ^B'A'D'  =  4  C'A'D',  therefore, 

by  III  5,  *  BAD"  =  *  CAD".       Since  further 

4  BAD  is  congruent  to  ^CVIZ},  so  there  is  (by  48) 
within    the    angle    CAD    a    ray    AD'"    such    that 


Df    °    ,D* 


C 

Fig.  si, 


$BAD"=$CAD'"  and  also  ^fDAD"^  4  DAD'". 
But  we  had  ^BAD"=  4  CAD",  and  therefore  we 
must  (by  III  5)  also  have  t{.CAD"=  if  CAD'". 
This  is  impossible,  since  (by  III  4)  every  angle  can 
be  set  off  against  a  given  ray  toward  a  given  side  in 
a  given  plane  only  in  one  way. 

Herewith  is  the  proof  for  the  congruence  of  right 
angles  completed. 

52.  Corollary  to  51.  At  a  point  A  of  a  straight  a 
there  is  not  more  than  one  perpendicular  to  a. 

53.  Definition.  When  any  two  angles  are  con- 
gruent to  two  adjacent  angles,  each  is  said  to  be  the 
supplement  of  the  other. 

54.  Definition.  If  any  angle  can  be  set  off  against 
one  of  the  rays  of  a  right  angle  so  that  its  second 


28 


RATIONAL    GEOMETRY. 


side  lies  within  the  right  angle,  it  is  called  an  acute 
angle. 

55.  Definition.  Any  angle  neither  right  nor  acute 
is  called  an  obtuse  angle. 

56.  Definition.  A  triangle  with  two  sides  con- 
gruent is  called  an  isosceles  triangle. 

57.  Theorem.  The  angles  opposite  the  congruent 
sides  of  an  isosceles  triangle  are  congruent. 

Let  ABC  be  an  isosceles  triangle, 
having  AB^BC. 

To  prove  ^  A  ==  if.C. 

Proof.  Since  in  the  triangles 
ABC  and  CBA  we  have  the  con- 
gruences AB  =  CB,  BC  =  BA, 
^ABC=^f  CBA ,  therefore  (by 
III  6)    $CAB=ifACB. 

58.  (Third  congruence  theorem  for  triangles.) 
Two  triangles  are  congruent  if  the  three  sides  of  the 
one  are  congruent,  respectively,  to  the  three  sides  of  the 
other. 


Fig. 


Fig.  23. 


In  the  triangles  ABC  and  A'B'C  take  ABsA'B', 
ACmA'C,  BCmB'C. 


CONGRUENCE.  29 

To  prove  a  ABC  =  a  A'B'C. 

Proof.  In  the  plane  of  ABC  toward  the  side  of 
the  straight  AC  not  containing  B  against  the  ray 
AC  take  the  angle  CAB"=  CA'B'.  Take  the 
sect  AB"  =  A' B'.  Then  (by  43)  &AB"C  =  *A'B'C. 
Therefore  B"C  =  BC,  and  A  BCB"  is  isosceles ;  there- 
fore (by  5  7)  7f  CBB"  =  t(.  CB"B.  So  also  is  a  BAB" 
isosceles  and  ,\  *  ABB"  =  if  AB"B.  Therefore 
(by  49)  the  angle  ABC=^AB"C.  But  ^AB"C 
=  4.  A'B'C.     :.  (by  43)   aABC^  a  A'B'C. 

59.  If  A,  B,  C  be  any  three  points  not  costraight, 
then  (by  the  method  used  in  58)  we  can  construct 
a  point  B"   such  that  AB"  =  AB  and  CB"  =  CB. 

Therefore  a  point  D  such  that  no  other  point 
whatsoever,  say  D",  gives  AD"  =  AD  and  CD"  = 
CD,  must  be  costraight  with  AC. 

The  following  have  been  given  as  definitions: 

If  A  and  B  are  two  distinct  points,  the  straight 
AB  is  the  aggregate  of  points  P  for  none  of  which 
is  there  any  point  Q  such  that  QA  a  PA  and  QB  = 
PB. 

If  A,  B,  C  are  distinct  points  not  costraight, 
the  plane  ABC  is  the  aggregate  of  points  P  for 
none  of  which  is  there  any  point  Q  such  that 
QA=PAt  QB^PB,  and  QC^PC. 

60.  Convention.  Any  finite  number  of  points  is 
called  a  figure;  if  all  points  of  the  figure  lie  in  3 
plane,  it  is  called  a  plane  figure. 

61.  Convention.  Two  figures  are  called  congruent 
if  their  points  can  be  so  mated  that  the  sects  and 
angles  in  this  way  coupled  are  all  congruent. 

Congruent  figures  have  the  following  properties: 


3©  RATIONAL   GEOMETRY. 

If  three  points  be  costraight  in  any  one  figure  their 
mated  points  are  also,  in  every  congruent  figure, 
costraight.  The  distribution  of  points  in  corre- 
sponding planes  in  relation  to  corresponding  straights 
is  in  congruent  figures  the  same ;  the  like  holds  for 
the  order  of  succession  of  corresponding  points  in 
corresponding  straights. 

62.  The  most  general  theorem  of  congruence  for' 
the  plane  and  in  general  is  expressed  as  follows : 

If  (A,  B,  C,  .  .  .)  and  (A't  B',  C,  .  .  .)  are  con- 
gruent plane  figures  and  P  denotes  a  point  in  the 
plane  of  the  first,  then  we  can  always  find  in  the 
plane  of  the  second  figure  a  point  P'  such  that 
(A,  B,C,  .  .  .  ,  P)  and  (A',  B\  C,  .  .  .  ,  P')  are  again 
congruent  figures. 

If  each  of  the  figures  contains  at  least  three  non- 
cost  raight  points,  then  is  the  construction  of  P'  only 
possible  in  one  way. 

If  (A}  B,  C,  .  .  .)  and  (A',  B',  C ,  .  .  .)  are  con- 
gruent figures  and  P  any  point  whatsoever,  then  we 
can  always  find  a  point  P't  such  that  the  figures  (A, 
B,  C,  .  . .  ,  P)  and  (A',  B\  C,  .  .  .  ,  Pf)  are  congruent. 

If  the  figure  {A,  B,  C,  .  .  .)  contains  at  least  four 
non-coplanar  points,  then  the  construction  of  P'  is 
only  possible  in  one  way. 

This  theorem  contains  the  weighty  result,  that 
all  facts  of  congruence  are  exclusively  conse- 
quences (in  association  with  the  assumption-groups 
I  and  II)  of  the  six  assumptions  of  congruence 
already  above  set  forth. 

This  theorem  expresses  the  existence  of  a  cer- 
tain reversible  unique  transformation  of  the  aggre- 


CONGRUENCE.  31 

gate  of  all  points  into  itself  with  which  we  are 
familiar  under  the  name  of  motion  or  displacement. 

We  have. here  founded  the  idea  of  motion  upon 
the  congruence  assumptions.  Thereby  we  have 
based  the  idea  of  motion  on  the  congruence  idea. 

The  inverse  way,  to  try  to  prove  the  congruence 
assumptions  and  theorems  with  help  of  the  motion 
idea,  is  false  and  fallacious,  since  the  intuition  of 
rigid  motion  involves,  contains,  and  uses  the  con- 
gruence idea. 

63.  Exercises. 

Ex.  6.  Show  a  number  of  cases  where  two  straights 
determine  a  point.  Show  cases  where  two  straights  do 
not  determine  a  point.  Are  any  of  these  latter  pairs 
coplanar? 

Ex.  7.  Show  cases  where  three  coplanar  straights  deter- 
mine 3  points;  2  points;  1  point.  Are  there  cases  where 
they  determine  no  point? 

Ex.  8.  How  many  straights  are,  in  general,  deter- 
mined by  3  points?  by  4  coplanar  points?  What  special 
cases  occur? 

Ex.  9.  Any  part  of  a  triangle  together  with  the  two 
adjoining  parts  determine  the  3  other  parts.     Explain. 

Ex.  10.  Try  to  state  the  first  two  congruence  theorems 
for  triangles  so  that  either  can  be  obtained  from  the  other 
by  simply  interchanging  the  words  side  and  angle. 

Ex.  11.  Principle  of  Duality  in  the  Plane. 

In  theorems  of  configuration  and  determination  we 
may  interchange  point  and  straight,  sect  and  angle. 
Try  to  write  down  a  theorem  of  which  the  dual  is  true; 
is  false. 

Ex.  12.  If  two  angles  of  a  triangle  are  congruent  it  is 
isosceles. 

Ex.  13.  If  the  sides  of  a  A  are  ^,  so  are  the  a£s. 
Dual? 


32  RATIONAL   GEOMETRY. 

Ex.  14.  In  an  isosceles  A,  sects  to  the  sides  from  the 
ends  of  the  base  making  with  it    m  sfs  are    = . 

Ex.  15.  If  any  two  sects  from  the  ends  of  a  side  of  a 
A  to  the  other  sides  making  =  ^s  are  =,  the  A  is  isosceles. 

64.  Definition.  Two  parallels  are  coplanar 
straights  with  no  common  point. 

65.  No  assumption  about  parallels  is  necessary 
for  the  establishment  of  the  facts  of  congruence  or 
motion. 

66.  Theorem.  Through  a  point  A  without  a 
straight  a  there  is  always  one  parallel  to  a. 

Proof.  Take  the  ray  from  the  given  point  A 
through  any  point  B  of  the  straight  a.  Let  C  be 
any  other  point  of  the  straight  a.  Then  take  in  the 
plane  ABC  an  angle  congruent  to  ^fABC  against 
AB  at  the  point  A  toward  that  side  not  containing 
(7.     The  straight  so  obtained  through  A  does  not 


^L 


B  C  D 

Fig.  24. 
meet  straight  a.  If  we  supposed  it  to  cut  a  in  the 
point  D,  and  that,  say,  B  lay  between  C  and  D, 
then  we  could  take  on  a  a  point  D' ',  such  that  B  lay 
between  D  and  D',  and  moreover  AD  =  BD' .  Be- 
cause of  the  congruence  of  the  triangles  ABD  and 
BAD'  (by  43),  therefore  4 ABD  =  ^BAU \  and 
since  the  angles  ABD'  and  ABD  are  adjacent 
angles,  so  must  then,  having  regard  to  45,  also  the 
angles  BAD  and  BAD'  be  adjacent  angles.  But 
because  of  6,  this  is  not  the  case. 


CONGRUENCE.  S3 

67.  Definition.  A  straight  cutting  across  other 
straights  is  called  a  transversal. 

68.  Definition.  If,  in  a  plane,  two  straights  are 
cut  in  two  distinct  points  A,  B  by  a  transversal,  at 
each  of  these  points  four  angles  are  made.  Of 
these  eight,  four,  having  each  the  sect  A B  on  a  side 
[e.g.,  3,  4,  1',  2'],  are  called  interior  angles.  The 
other  four  are  called  exterior  angles.  Pairs  of  angles, 
one  at  each  point,  which  lie 
on  the  same  side  of  the 
transversal,  the  one  exterior 
and  the  other  interior,  are 
called  corresponding  angles 
[e.g.,  1  and  1']. 

Two  non-adjacent  angles 
on    opposite    sides  of    the  IG'  25* 

transversal,  and  both  interior  or  both  exterior,  are 
called  alternate  angles  [e.g.,  3  and  1']. 

Two  angles  on  the  same  side  of  the  transversal, 
and  both  interior  or  both  exterior,  are  called  con- 
jugate angles  [e.g.,  4  and  i']. 

69.  Theorem.  Two  coplanar  straights  are  parallel 
if  a  transversal  makes  congruent  alternate  angles. 
[Proved  in  66.] 

70.  Theorem.  If  two  straights  cut  by  a  trans- 
versal have  corresponding  angles  congruent  they  are 
parallel. 

Proof.  The  angle  vertical  to  one  is  alternate  to 
the  other. 

Ex.  16.  If  two  corresponding  or  two  alternate  angles 
are  congruent,  or  if  two  interior  or  two  exterior  angles 
on   the   same   side   of   the   transversal   are   supplemental, 


34  RATIONAL    GEOMETRY. 

then  every  angle  is  congruent  to  its  corresponding  and 
to  its  alternate  angle,  and  is  supplemental  to  the  angle 
on  the  same  side  of  the  transversal  which  is  interior  or 
exterior  according  as  the  first  is  interior  or  exterior. 

Ex.  17.  If  two  interior  or  two  exterior  angles  on  the 
same  side  of  the  transversal  are  supplemental,  the  straights 
are  parallel. 

Ex.  18.  Two  straights  perpendicular  to  the  same 
straight  are  parallel. 

Ex.  19.  Construct   a  right   angle. 

Ex.  20.  On  the  ray  from  the  vertex  of  a  triangle  co- 
straight  with  a  side  take  a  sect  congruent  to  that  side. 
The  two  new  end-points  determine  a  straight  parallel  to  the 
triangle's  third  side. 

Ex.  21.  On  one  side  of  any  if  with  vertex  A  take  any 
two  sects  AB,  AC  and  on  the  other  side  take  congruent 
to  these  AB' ,  AC .  Prove  that  BC  and  B'C  intersect,  say 
at  D.  Prove  BC'^B'C,  aBCD^aB'CD,  if  BAD  = 
ifB'AD. 

Ex.  22.  From  two  given  points  on  the  same  side  of  a 
given  st'  find  st's  crossing  on  that  given  st\  and  making 
congruent   if  's  with  it. 

Ex.  23.  Construct  a  triangle,  given  the  base,  an  angle 
at  the  base,  and  the  sum  of  the  other  two  sides  [A  from 
a,  b,  a-\-c]. 

Ex.  24.  If  the  pairs  of  sides  of  a  quadrilateral  not  con- 
secutive are  congruent,  they  are  ||. 

Ex.  25.  On  a  given  sect  as  base  construct  an  isosceles  a. 

Ex.  26.  If  on  the  sides  AB,  BC,  CA  of  an  equilateral 
a,  AD  =BE=CF,  then  ADEF  is  equilateral,  as  is  A 
made  by  AE,  BF,  CD. 


CHAPTER  IV. 

PARALLELS. 
IV.  Assumption  of  Parallels  (Euclid's  Postulate). 

IV.  Through  a  given  point  there  is  not  more  than 
one  parallel  to  a  given  straight. 

71.  The  introduction  of  this  assumption  greatly 
simplifies  the  foundation  and  facilitates  the  con- 
struction of  geometry. 

72.  Theorem.  Two  straights  parallel  to  a  third  arc 
parallel. 

Proof.  Were  1  and  2  not  parallel,  then  there 
would  be  through  their  intersection  point  two  par- 
allels to  3,  which  is  in  contradiction  to  IV. 

73.  Theorem.  //  a  transversal  cuts  two  parallels, 
the  alternate  angles  are  congruent. 

Proof.  Were  say  ^ BAD  not  =  ^ABC,  then  we 
could  through  A  (by  III  4)  take  a  straight  making 
^.BAU^i-ABC    [D'  and  / 

D  on  same  side  of  AB],  and   _d a/        «' 

so  we  would  have  (by  69) 
through  A  two  parallels  to 
a,  in  contradiction  to  IV.       

74.  Corollary   to    73.     A 
perpendicular  to  one  of  two  Fig.  26. 
parallels  is  perpendicular  to  the  other  also. 

35 


36  RATIONAL   GEOMETRY. 

\ 

75.  Theorem.     If  a  transversal  cuts  two  parallels, 

the  corresponding  angles  are  congruent. 

Proof.  The  angle  vertical  to  one  is  alternate  to 
the  other. 

Ex.  27.  A  straight  meeting  one  of  two  parallels  meets 
the  other  also. 

Ex.  28.  A  straight  cutting  two  parallels  makes  con- 
jugate angles  supplemental. 

Ex.  29.  If  alternate  or  corresponding  angles  are  un- 
equal or  if  conjugate  angles  t  are  not  supplemental,  then 
the  straights  meet.     On  which  side  of  the  transversal? 

76.  Theorem.  A  perpendicular  to  one  of  two 
parallels  is  parallel  to  a  perpendicular  to  the  other. 

Proof.  Either  of  the  two  given  parallels  makes 
(by  74)  right  angles  with  both  perpendiculars,  which 
therefore  are  parallel  by  69. 

77.  Corollary  to  76.  Two  straights  respectively 
perpendicular  to  two  intersecting  straights  cannot 
be  parallel. 

Proof.  For  if  they  were  parallel,  then  (by  76) 
the  intersecting  straights  would  also  be  parallel. 

78.  Convention.  When  two  angles  are  set  off 
from  the  vertex  of  a  third  against  its  sides  so  that  no 
point  is  interior  to  two,  if  the  two  sides  not  common 
are  costraight,  the  three  angles  are  said  together  to 
form  two  right  angles. 

79.  The  angles  of  a  triangle  together  form  two  right 

angles. 

Proof.     Take    alternate 

tCBF=*C  and  $ABD 
m  gA ;  then  (by  69)  can 
neither  BF  nor  BD  cut  AC. 
By  the  parallel  postulate 
IV,  then  is  FBD  a  straight. 


Fig. 


PARALLELS. 


37 


80.  Theorem.  If  two  angles  of  one  triangle  are 
congruent  to  two  of  another,  then  the  third  angles 
are  congruent. 

Proof.  Given  tA=^Af  and  4-B^4.B'.  Take 
CP  parallel  (||)  to  AB  and  C'P'  ||  to  A'B'.     Then 


ta^i-A  and  ^.p  =  ^By  ^.a'^^A'  and  ^/?'  = 
*B'.  .'.(by  49)  *BCD^*B'C'D'.  /.(by  45) 
the  adjacent  angles  Z  ACB  =  $  A'C'B'. 

81.  Theorem.  Two  triangles  are  congruent  if  they 
have  a  side,  an  adjoining  and  the  opposite  angle  re- 
spectively congruent. 

Proof.     By  80  and  44. 

Ex.  30.  Every  triangle  has  at  least  two  acute  angles. 

Ex.  31.  If  the  rays  of  one  angle  are  parallel  or  per- 
pendicular to  chose  of  another,  the  angles  are  congruent 
or  supplemental. 

Ex.  32.  In  a  right-angled  triangle  [a  triangle  one  of 
whose  angles  is  a  right  angle]  the  two  acute  angles  are 
complemental  (calling  two  angles  complements  which 
together  form  a  right  angle). 

82.  Theorem.  In  any  sect  AB  there  is  always  one 
and  only  one  point  C  such  that  AC  =  BC. 

Proof.  Take  any  angle  BAD  at  A  against  AB, 
and  the  angle  congruent  to  it  at  B  against  BA  and 
on  the  opposite  side  of  a  in  the  plane  BAD]  and 
take  any  sect  AD  on  the  free  ray  from  A,  and  one 


38  RATIONAL   GEOMETRY. 

BF  congruent  to  it  on  the  free  ray  from  B.  The 
sect  DF  must  cut  a,  say  in  C,  since  D  and  F  are  on 
opposite  sides  of  a.  Moreover,  C  is  between  A  and 
B.  Otherwise  one  of  them,  say  A ,  would  be  between 
B  and  C.     But  then  DA  would  have  a  point  A  on 


Fig.  29. 

BC,  a  side  of  triangle  FBC,  and  so  (by  II  4)  must 
meet  another  side.  But  this  is  impossible,  since  it 
meets  FC  produced  at  D  and  is  parallel  to  BF. 
Since  thus  ^A=^B,  and  i-ACD^^BCp  [ver- 
tical], therefore  (by  81)   aACD=aBCF. 

Therefore  AC  =  BC. 

If  we  suppose  a  second  such  point  C,  then  on 
ray  DC  take  CF'  =  DC.  Therefore  (by  43) 
i  CBFf  =^DAC=4ABF,  and  BF'  =  AD  =  BF. 
Therefore  Ff  is  F  and  C  is  C. 

83.  Convention.  The  point  C  of  the  sect  A B  such 
that  i4C  =  BC  may  be  called  the  bisection-point  of 
A  J5,  and  to  fo*s£c/  A  5  shall  mean  to  take  this  point  C. 

Ex.  ^t,.  Parallels  through  the  end-points  of  a  sect 
intercept  congruent  sects  on  any  straight  through  its 
bisection-point. 

Ex.  34.  In  a  right-angled  triangle  the  bisection-point 
of  the  hypothenuse  (the  side  opposite  the  r't^!)  makes 
equal  sects  with  the  three  vertices. 

Hint.     Take  one  acute  2^  in  the  r't  ^. 


PARALLELS.  39 

84.  Theorem.  Within  any  •£  (h,  k)  there  is  always 
one  and  only  one  ray,  /,  such  that  •£  (h,  /)  =  ^  (k,  I). 

Proof.     From  the  vertex  0  take  OA  =  OA'.     By 
82  take  C,  the  bisection-point 
of   A  A'.       Then    aAOC=a 
A'OC     [As   with     3    sides  = 
are=].     .\  ^AOC=^fA'OC. 

If  we  suppose  a  second  such 
ray  OC,  then  a  ,4  0(7=  a 
A'OC     [as     with     2     sides  FlG 

and  the  included    a£  =are  =  ]. 
i.ACmA'C    :.  (by  82)  CisC. 

85.  Convention.  The  ray  /of  ?{  (h,  k)  such  that 
^  (/t,  /)  =  ^  (/,  &)  may  be  called  the  bisection  ray  or 
bisector  of  ^  (/t,  &),  and  to  bisect  a£  (/*,  &)  shall  mean 
to  take  this  ray  /. 

Ex.  35.  An  angle  may  be  separated  into  2,  4,  8,  1 6,  ...  v 
2M  congruent  angles. 

Symmetry. 

86.  Definition.     Two  points  are  said  to  be  sym- 
metrical with  regard  to  a  straight,  when  it  bisects 

at  right  angles  their  sect.  The  straight  is 
called  their  axis  of  symmetry.  Two  points 
have  always  one,  and  only  one,  symmetry 
axis. 

A  point  has,  with  regard  to  a  given  axis 

of  symmetry,  always  one,  and  only  one,  sym- 

Fig.  31.  metrical  point,  namely,  the  one  which  ends 

the  sect  from  the  given  point  perpendicular  to  the 

axis  and  bisected  by  the  axis. 


4o 


RATIONAL    GEOMETRY. 


Z, 


Fig.  S3. 


87.  Definition.  Two  figures 
have  an  axis  of  symmetry  when, 
with  regard  to  this  straight, 
every  point  of  each  has  its  sym- 
metrical point  on  the  other. 
One  figure  has  an 
Fig.  32.  axis    of    symmetry 

when,  with  regard  to  this  straight,  every 
point  of  the  figure  has  its  symmetrical 
point  on  the  figure. 

One  figure  is  called  symmetrical  when 
it  has  an  axis  of  symmetry. 

Any  figure  has,  with  regard  to  any  given  straight 
as  axis,  always  one,  and  only  one,  symmetrical  figure. 
88.  Theorem.     An  angle  is  symmetrical  with  re- 
gard to  its    bisector    and    the    end-points 
of   congruent  sects  from   the  vertex    are 
symmetrical. 

Proof.     Their  sect  is  bisected  at  right 
angles  by  the  angle-bisector. 

89.  Definition.  A  sym- 
metrical quadrilateral 
with  a  diagonal  as  axis  is  called  a 
deltoid. 

90.  Definition. 
A  sect  whose  end- 
points  are  the  bi- 
section-points of  opposite  sides 
of  a  quadrilateral  is  called 
a  median.  So  is  the  sect  from  a 
vertex  of  a  triangle  to  the  bisec- 
tion-point of  the  opposite  side. 


Fig.  34. 


Fig.  35. 


PARALLELS.  41 

91.  Definition.  A  symmetrical  quadrilateral  with 
a  median  as  axis  is  called  a  symira. 

Ex.  36.  In  a  r't  A  if  to  set  off  one  acute  £,  a,  in  the 
other,  /?,  bisects,  so  is  it  with  /?'s  sides. 

Ex.  37.  The  st'  through  the  bisection-point  of  the  base 
of  a  I-  A ,  and  the  opposite  vertex  is  ±  to  the  base  and 
bisects    if . 

Ex.  38.  The  r't  bi'  of  the  base  of  a  +  A  bisects  if  at  the 
vertex. 

Ex.  39.  The   _L    from  vertex  bisects    base    and   ^   in  a 

•I-  A. 

Ex.  40.  The  bisector  of  ^  at  vertex  of  a  -I-  A  is  r't 
bi'  of  base. 

Ex.  41.  If  a  r't  bi'  of  a  side  contains  a  vertex,  the  A 
is  -I-. 

Ex.  42.  The  bisector  of  an  exterior  ^  at  vertex  of  +  a 
is  ||  to  base,  and  inversely. 

Ex.  43.  The  end  of  sect  from  intersection  of  congruent 
sides  of  a  +  A  costraight  and  ■  to  one  determines  with 
end  of  other  a  ±  to  base. 

Ex.  44.  To  erect  a  i.  at  the  end-point  of  a  sect  without 
producing  the  sect. 

Ex.  45.  A  ||  to  one  side  of  an  ^  makes  with  its  bisector 
and  other  side  a  -I-  A. 

Ex.  46.  The  bisectors  of  the  ■  ?f  s  of  a-l-  A  are  =. 

Ex.  47.  Every  symmetrical  quadrilateral  not  a  del- 
toid is  a  symtra. 

Ex.  48.  The  intersection  point  of  two  symmetrical 
straights  is  on  the  axis. 

Ex.  49.  The  bisector  of  an  angle  is  symmetrical  to  the 
bisector  of  the  symmetrical  angle. 

Ex.  50.  A  figure  made  up  of  a  straight  and  a  point  is 
symmetrical. 

Ex.  51.  In  any  deltoid  [1]  One  diagonal  (the  axis) 
is  the  perpendicular  bisector  of  the  other.  [2]  One 
diagonal  (the  axis)  bisects  the  angles  at  its  two  vertices. 
[3]    Sides  which  meet  on  one  diagonal  (the  axis)  are  con- 


42  RATIONAL    GEOMETRY. 

gruent;  so  each  side  equals  one  of  its  adjacent  sides.  [4] 
One  diagonal  (not  the  axis)  joins  the  vertices  of  congruent 
angles  and  makes  congruent  angles  with  the  congruent 
sides.  [5]  The  triangles  made  by  one  diagonal  (the  axis) 
are  congruent.  [6]  One  diagonal  (not  the  axis)  makes 
two  isosceles  triangles. 

Ex.  52.  Any  quadrilateral  which  has  one  of  the  six 
preceding  pairs  of  properties  (Ex.  51)  is  a  deltoid. 

Ex.  53.  A  quadrilateral  with  a  diagonal  which  bisects 
the  angle  made  by  two  sides,  and  is  less  than  each  of  the 
other  two  sides,  and  these  sides  congruent,  is  a  deltoid 
with  this  diagonal  as  axis. 

Ex.  54.  A  quadrilateral  with  a  side  meeting  a  con- 
gruent side  in  a  greater  diagonal  which  is  opposite  con- 
gruent angles  is  a  deltoid  with  that  diagonal  as  axis. 

Ex.  55.  In  any  symtra  [1]  Two  opposite  sides  are 
parallel,  and  have  a  common  perpendicular  bisector. 
[2]  The  other  two  sides  are  congruent  and  make  con- 
gruent angles  with  the  parallel  sides. 

[3]  Each  angle  is  congruent  to  one  and  supplemental 
to  the  other  of  the  two  not  opposite  it. 

[4]  The  diagonals  are  congruent  and  their  parts  adja- 
cent to  the  same  parallel  are  congruent. 

[5]  One  median  bisects  the  angle  between  the  two 
diagonals,  and  also  the  angle  between  the  non-parallel 
sides  (produced). 

Ex.  56.  Any  quadrilateral  which  has  one  of  the  pre- 
ceding five  pairs  of  properties  (Ex.  55)  is  a  symtra. 

92.  Definition.  A  trapezoid  is  a  quadrilateral 
with  two  sides  parallel. 

93.  Definition.  A  parallelogram  is  a  quadrilateral 
with  each  side  parallel  to  another  (its  opposite). 

94.  Definition.  A  parallelogram  with  one  angle 
right  is  called  a  rectangle.  A  parallelogram  with 
two  consecutive  sides  congruent  is  called  a  rhombus. 
A  rectangle  which  is  a  rhombus  is  called  a  square. 


PARALLELS.  43 

95.  Theorem.     The  opposite  sides  and  angles  of  a 


Fig.  37. 

parallelogram  are  congruent,  and  its  diagonals  bisect 
each  other. 

Proof.  aABC=  a  ADC  [side  and  2  adjoining 
*ss].  :.BC  =  AD.  .'.  (as  in  82)  AF  =  FC  and 
BF  =  FD. 

96.  Theorem.  //  three  parallels  make  congruent 
sects  on  one  transversal,  they  do  on  every  transversal. 

Given  a||6||c,  also  AB  = 
BC. 

To  prove  FG  =  GH. 

Proof.  Take  FL\\GM\\ 
AB.  Then  FL^AB^BC 
=  GM  [95,  opposite  sides  of 
a||gm  are  =  ].  .*.  aFLG  = 
aGMH  [side  and  2  adjoining  i-s  =  ].     .'.  FG  =  GH. 

97.  Corollary  to  96.  A  straight  through  the  bi- 
section-point of  one  side  of  a  triangle  and  parallel  to 
a  second  side  bisects  the  third  side.  [In  figure  let 
F  coincide  with  A.] 

98.  Inverse  of  97.  The  straight  through  the  bi- 
section-points of  any  two  sides  of  a  triangle  is 
parallel  to  the  third  side.  [For,  by  97,  it  is  iden- 
tical with  the  ||  to  the  third  side  through  either 
bisection- point.] 


Fig.  38. 


44 


RATIONAL    GEOMETRY. 


99.  Theorem.  The  sect  whose  end- points  are  the 
bisection- points  of  two  sides  of 
a  triangle  is  congruent  to  each 
sect  made  in  bisecting  the  third 
side. 

Proof.     By  97  GH  \\  BC  bi- 
sects   AC.     Since     (by    98) 
Fig.  39.  FG\\CH,  :.  (by  95)  FG^CH. 

100.  Theorem.     If  two  sides  of  a  quadrilateral  ore 
congruent  and  parallel  it  is  a  parallelogram. 

Given  AB  =  and  \\CD. 
Proof,     a  ABC  a  a  ADC. 


Fig.  40. 

/.  ^fACB=  if  CAD.     :.  CB\\AD. 

Ex.  57.  Every  straight  through  the  intersection  of 
its  diagonals  cuts  any  parallelogram  into  congruent 
trapezoids. 

Ex.  58.  A  quadrilateral  with  each  side  equal  to  its 
opposite  is  a  parallelogram. 

Ex.  59.  A  quadrilateral  with  a  pair  of  opposite  sides 
equal,  and  each  greater  than  a  diagonal,  making  equal 
alternate  angles  with  the  other  sides,  is  a  parallelogram. 

Ex.  60.  A  quadrilateral  with  a  side  equal  to  its  oppo- 
site, and  less  than  a  diagonal  opposite  equal  angles,  is  a 
parallelogram . 

Ex.  61.  A  quadrilateral  with  each  angle  equal  to  its 
opposite  is  a  parallelogram. 

Ex.  62.  A  quadrilateral  whose  diagonals  bisect  each 
other  is  a  parallelogram. 


PARALLELS. 


45 


10 1.  Theorem.  In  any  sect  AB  there  are  always 
two,  and  only  two,  points,  C,  D,  such  that  AC  = 
CD^DB. 


Fig.  41. 

Proof.  Take  on  any  ray  from  A,  any  sect  AF, 
and  a  sect  FG  =  AF,  and  a  sect  GH  =  FG.  Take 
FC\\GD\\HB.     Then,  by  96,  AC^CD^DB. 

Suppose  two  other  such  points  C \  D' '.  Then,  by 
98,  C'F  ||  D'G.  Now  HB'  \\  GD'  (by  96)  makes  D'B' 
s-D'C.  .'.  from  our  hypothesis  and  III  1,  B'  is 
identical  with  B.  .'.  since  GD\\HB  (by  IV)  D'  is 
identical  with  D.  .'.  since  FCJGT?  (by  IV)  C  is 
identical  with  C. 

102.  The  two  points,  C,  D,  of  the  sect  AB  such 
that  AG  =  CD  =  DB  may  be  called  the  trisection- 
points  oi  AB. 

103.  Theorem.  77t£  three  medians  of  a  triangle  are 
copunctal  in  that  trisection-point  of  each  remote  from 
its  vertex. 

Proof.  Any  median  AG  must  meet  any  other 
CF}  since  A  and  G  are  on  dif- 
ferent sides  of  the  straight  CF, 
and  so  the  cross  of  st'  AG 
with  st'  CF  is  on  sect  AG,  and 
similarly  it  is  on  sect  CF.  If 
P,  Q,  are  bisection-points  of 
OC  and  OA,  then  (by  98  and  Fig.  42. 

99)  PQ  ||  and  =GF.     .'.  by  100  PQFG  is  a  ||  gm  and 
(by  95)  PF  and  QG  bisect  each  other. 


46  RATIONAL   GEOMETRY. 

104.  Definition.  The  cointersection-point  of  its 
medians  is  called  the  triangle's  centroid. 

105.  Definition.  A  perpendicular  from  a  vertex  to 
the  straight  of  the  opposite  side  is  called  an  altitude 
of  the  triangle.  This  opposite  side  is  then  called 
the  base.  The  perpendicular  from  a  vertex  of  a 
parallelogram  to  the  straight  of  a  side  not  through 
this  vertex  is  called  the  altitude  of  the  parallelogram 
with  reference  to  this  side  as  base. 

Ex.  63.  The  bisectors  of  the  four  angles  which  two 
intersecting  straights  make  with  each  other  form  two 
straights  perpendicular  to  each  other. 

Ex.  64.  If  four  coinitial  rays  make  the  first  angle  con- 
gruent to  the  third,  and  the  second  congruent  to  the 
fourth,  they  form  two  straights. 

Ex.  65.  How  many  congruent  sects  from  a  given  point 
to  a  given  straight? 

Ex.  66.  Does  the  bisector  of  an  angle  of  a  triangle 
bisect  the  opposite  side? 

Ex.  67.  The  bisectors  of  vertical  angles  are  costraight. 

Ex.  68.  If  two  isosceles  triangles  be  on  the  same  base 
the  straight  determined  by  their  vertices  bisects  the 
base  at  right  angles. 

Ex.  69.  Suppose  a  A  to  be  3  bars  freely  jointed  at  the 
vertices.  Is  it  rigid?  Are  the  ^s  fixed  and  the  joints 
of  no  avail?  Of  what  theorem  is  this  a  consequence? 
How  is  it  with  a  jointed  quadrilateral?     Why? 

Ex.  70.  Joining  the  bisection-points  of  the  sides  of  a  A 
cuts  it  into  4  =  As. 

Ex.  71.  Joining  the  bisection-points  of  the  consecutive 
sides  of  a  quadrilateral  makes  a  ||  g'm. 

Ex.  72.  The  medians  of  a  quadrilateral  and  the  sect 
joining  the  bisection-points  of  its  diagonals  are  all  three 
bisected  by  the  same  point. 

Ex.  73.  If   the   bisection-points    of   two    opposite    sides 


PARALLELS.  47 

of  a  ||  g'm  are  joined  to  the  vertices  the  diagonals  are  tri- 
sected. 

Ex.  74.  The  J_s  from  any  point   in  the  base  of   an   f  A 
to  the  sides  are  together  an  altitude. 

Ex.  75.  The  diagonals  of  a  rectangle  are    =,  of  a  rhom- 
bus are  ±. 

Ex.  76.  If  2  ||  s  are  cut  by  a  transversal,  the  bisectors 
of  the  interior  ^  s  make  a  rectangle. 

Ex.  77.  The  angle-bisectors  of  a  rectangle  make  a  square. 
Ex.  78.   If  the  ^  s  adjoining  one  of  the  ||  sides  of  a  trape- 
zoid are   ■,  so  are  the  others. 

Ex.  79.  The, bisectors  of  the  interior  ^s  of  a  trapezoid 
make  a  quad'  with  2  r't  ^s. 

Ex.  80.  The    bisection-point    of    one    sect    between    ||s 
bisects  any  through  it. 

Ex.  81.  The     =    altitudes   in  -I-  A  make    with    the    base 
if  s  ss  to  those  made  in  bisecting  the  other  ^  . 

Ex.  82.  Through   a   given   point   within   an    ^    draw   a 
sect  terminated  by  the  sides  and  bisected  by  the  point. 

Ex.  83.  Sects  from  the  vertex  to  the  trisection-points 
of  the  base  of  +  A  are    =. 

Ex.  84.  If  the  ^  s  made  by  producing  a  side  of  a  A  are 
s,  so  are  the  other  sides. 

Ex.  85.  If  a  quad'  has  2  pairs  of  congruent  consecutive 
sides,  the  other  ^s  are   =. 

Ex.  86.  Two  As"  are  =  if  two  sides  and  one's  median  are 
respectively  = . 

Ex.  87.  Two  +  As  are   =  if  one  ^  and  altitude  are  ■  to 
the  corresponding. 

Ex.  88.  Two   as  are    =   if  a  side,  its  altitude  and  an 
adjoining  ^  are  respectively  = . 

Ex.  89.  If  2  altitudes  are   =  the  A  is  + . 
Ex.  90.  Two  As  are  =  if  two  sides  and  one's  altitude  are 
s  to  the  corresponding. 

Ex.  91.  Two  as  are   =  if  a  side,  its  altitude  and  median 
are  respectively  = . 

Ex.  92.  Two  as  are  =  if  a  side  and  the  other  2  altitudes 
are  respectively  ■ . 


48  RATIONAL   GEOMETRY. 

Ex.  93.  Two  as  are  m  if  a  side,  an  adjoining  ^  and 
its  bisector  are  respectively    =  . 

Ex.  94.  Two  equilateral  as  are   ■  if  an  altitude  is   =. 

Ex.  95.  The  bisector  is  within  the  if  made  by  altitude 
and  median. 

Ex.  96.  In  a  right  A  one  bisector  also  bisects  if  be- 
tween its  altitude  and  median. 

Ex.  97.  Two  sects  from  the  vertices  of  a  A  to  the  oppo- 
site sides  cannot  bisect  each  other. 

Ex.  98.  The  ±s  from  2  vertices  of  a  A  upon  the  median 
from  the  third  are    =. 

Ex.  99.  Two  ||  g'ms  having  an  if  and  the  including 
sides    =    are    =  . 

Ex.  100.  The  ±  from  the  circumcenter  to  a  side  is  half 
the  sect  from  the  opposite  vertex  to  the  orthocenter. 

Ex.  101.  The  centroid  is  the  trisection  point  of  the  sect 
from  orthocenter  to  circumcenter  remote  from  the  ortho- 
center 


CHAPTER  V. 

THE  CIRCLE. 

106.  Definition.  If  C  is  any  point  in  a  plane  a, 
then  the  aggregate  of  all  points  A  in  a,  for  which 
the  sects  CA  are  congruent  to  one  another,  is  called 
a  circle.     [qC(CA).] 

C  is  called  the  center  of  the  circle,  and  CA  the 
radius. 

107.  Theorem.  Any  ray  from  the  center  of  a 
circle  and  in  its  plane  a  cuts  the  circle  in  one,  and 
only  one,  point. 

108.  Theorem.  Any  straight  through  its  center 
and  in  its  plane  a  cuts  the  circle  in  two,  and  only 
two,  points,  and  these  are  on  opposite  sides  of  its 
center. 

Proof.  On  each  of  the  two  rays  determined  in 
this  straight  by  the  center  there  is  (by  III  1)  one, 
and  only  one,  sect  congruent  to  the  radius  of  the 
circle. 

109.  Definition.  A  sect  whose  end-points  are  on 
the  circle  is  called  a  chord. 

no.  Definition.  Any  chord  through  the  center 
is  called  a  diameter. 

in.  Theorem.  Every  diameter  is  bisected  by  the 
center  of  the  circle. 

49 


5° 


RATIONAL   GEOMETRY. 


Fig.  43- 


ii2.  Theorem.  No  circle  can  have  more  than  one 
center. 

Proof.  If  it  had  two,  the  diameter  through  them 
would  have  two  bisection -points,  which  (by  82)  is 
impossible. 

113.  Theorem.  The  straight  through  the  bisection- 
point  of  a  chord,  and  the  center  of  the  circle,  is  perpen- 
dicular to  the  chord. 

Proof.  A  ACO  =  A  BCO,  [  a  s  with 
3  sides  =  are  =];  .*.  if  ACO  = 
if  BCO.  But  they  are  adjacent; 
.'.  by  definition  COL  to  AB. 

114.  Corollary  to  113.  The  cir- 
cle is  symmetrical  with  regard 
to  any  one  of  its  diameters  as 
axis. 

115.  Corollary  to  1 1 3 .  If  with 
the  end-points  of  a  sect  each  of 
two  points  gives  congruent  sects 
the  two  determine  its  perpen- 
dicular bisector. 

116.  Corollary  to  1 1 5 .  If  two 
circles  have  two  points  in  com- 
mon their  center-straight  is  the 
perpendicular  bisector  of  their  .common  chord. 

117.  Theorem.  The  perpendicular  bisecting  any 
chord  contains  the  center.  The  perpendicular  from 
the  center  to  a  chord  bisects  it. 

Proof.  By  113  the  three  properties  pertain  to 
one  straight.  But  any  two  suffice  to  determine  that 
straight. 

118.  Corollary  to  117.     Every  point  which  taken 


THE   CIRCLE.  5 1 

with  two  points  gives  congruent  sects  is  on  the  per- 
pendicular bisector  of  their  sect. 

119.  Theorem.  Every  point  on  the  perpendicular 
bisector  of  a  sect  taken  with  its  end- points  gives  con- 
gruent sects. 

120.  Theorem.  A  straight  cannot  have  more  than 
two  points  in  common  with  a  circle. 

Proof.  If  it  had  a  third,  then,  "since  (by  117)  the 
perpendicular  bisecting  any  chord  contains  the  cen- 
ter, there  would  be  two  perpendiculars  from  the 
center  to  the  same  straight,  which  (by  47)  is  im- 
possible. 

121.  Theorem.  Chords  which  mutually  bisect  are 
diameters. 

Proof.  The  perpendicular  bisector  of  each  con- 
tains the  center. 

122.  Theorem.  Circles  with  three  points  in  com- 
mon are  identical. 

Proof.  The  center  is  on  the  perpendicular  bi- 
sectors of  the  chords. 

123.  Theorem.  Any  three  points  not  costraight  de- 
termine a  circle. 

Proof.  If  A ,  By  C  be  not  costraight,  bisect  (by  82) 
AB  at  D  and  BC  at  F  by 
perpendiculars.  [Take  (by 
III  4)  angles  =  to  ^C  in 
84.]  These  perpendiculars 
(by  77)  meet,  say  at  0. 
Therefore  (by  119)  A0  = 
BO  =  CO.     Therefore  A,  B,  Fig.  45- 

C  are  on  the  circle  with  center  0  and  radius  AO. 
By  122  the  three  are  on  no  other  circle. 


52 


RATIONAL   GEOMETRY. 


124.  Corollary  to  123  and  120.  [Points  on  the 
same  circle  are  called  concyclic.']  Every  three  points 
are  costraight  or  concyclic.  No  three  points  are 
costraight  and  concyclic. 

125.  Definition.  The  circle  through  the  vertices 
of  a  triangle  is  called  its  circumcircle,  O0(R),  and 
the  center  0  of  the  circumcircle  is  called  the  circum- 
center  of  the  triangle ;  its  radius,  R,  the  circum-radius. 

126.  Corollary  to  123.  The  three  perpendicular 
bisectors  of  the  sides  of  a  triangle  are  copunctal  in 
its  circumcenter. 

127.  Theorem.  The  three  altitudes  of  a  triangle  are 
copunctal. 

Given  the  a  ABC.  To  prove  that  the  straights 
through  A,  B,  C  perpendicular  to  the  straights  a,  b,  c 
respectively,  are  copunctal. 


Fig.  46. 

Proof      By  66,  through  A,  B,  C  take  B'C,  A'C, 

A'B'\BC,     AC,    AB    respectively.       /.  aAB'C  = 

aABC=  aABC  [as  with  a  side  and  2   adjoining 

^s=   are   =].     .'   AB'^AC,  and  AD  is  the  ±bi' 

of  B'C  [_L  to  1  st  of  2  ||  s  is  1  to  2nd]. 

Similarly  BEA.W  of  A'C'\  and  CFJ_bi'  of  A'B'. 
:.  AD,  BE,  CF  axe  copunctal  by  126. 


THE  CIRCLE 


53 


128.  Definition.  The  point  of  cointersection  of  the 
three  altitudes  is  called  the  orthocenter  of  the  triangle. 

129.  Theorem.  If  any  ray,  /,  be  taken  -within  a 
given  angle,  4  (K  &)>  the  bisectors  m,  n  of  the  two 
angles  so  made  form  an  angle  congruent  to  each  of 
the  angles  made  in  bisecting  the  given  angle. 

Proof.  On  the  other  side  of  h  from  k  take 
*(*,#)»*(*,*).  Then  since  *  (h,  n)  =  $  (n,  h) 
and  *(nffc)  s  *(*,#),    .\  (by  49)  *(M)^  *(»,#). 


Fig.  47. 

But  since  *  (n,  /)  ■  *  (fc,  «),  ,\  *  (n,  /)  =  ^  (£',  A). 
But  also  af  (/,  tn)  =  £(h,  m).  .*.  (by  49)  ^(w,  m)  = 
£(£',  w),  .*.  w  bisects  ^(w,  £').  .*.  (by  48)  £(w,  w) 
=  £(&,  /*)  where  6  bisects  £(/*,  fe). 

130.  Theorem.     The  bisectors  of  adjacent  angles 
make  a  right  angle. 

frtt 


Fig.  48. 
Proof.  Extend  one  of  the  bisectors,  as  /,  through 
the   vertex    0.      Then    *(/',  *)«*(/,  h')     [vertical 


54 


RATIONAL   GEOMETRY. 


^s  are    ■].      .*.  (by  III  5)    *(l',h)=t(l,k).     But 

by     hypothesis,     £(h,  m)  =  £(k,  m).        .'.(by     49) 
£(/,  m)  =  4-V ,  m).     .'.by  definition  £(l,  m)  is  right. 

Ex.  102.  If  a  straight  satisfy  any  two  of  the  following 
conditions  it  also  satisfies  the  others: 

1.  Passing  through  the  center. 

2.  Perpendicular  to  the  chord. 

3.  Bisecting  the  chord. 

4.  Bisecting  the  angle  at  the  center. 

Ex.  103.  Every  axis  of  symmetry  for  a  circle  contains 
the  center. 

Ex.  104.  Where  are  the  bisection- points  of  a  set  of 
parallel  chords? 

Ex.  105.  Where  are  the  bisection-points  of  a  set  of 
equal  chords? 

Ex.  106.  If  from  any  point  three  sects  drawn  to  a  circle 
are  congruent,  that  point  is  the  center. 

131.  Theorem.  If  any  ray  /  be  taken  without  a 
given  angle,  4-  (K  k) ,  and  of  the  two  angles  so 
formed,  one,  £(k,  I),  be  within  the  other,  if(k,  /), 
then  the  angle  formed  by  their  bisectors,  £(b,  n)t 
is  congruent  to  each  of  the  angles  as  4-  (k>  m)  made 
in  bisecting  the  given  angle. 


o 
Fig.  49. 

Proof.     By     129,     since    k    is    within     £(/,  h), 

.*.  tim,  n)  =4-(b,  I).     But  by  hypothesis   £(n,k)=- 

*(/,  »),  .*.  (by  49)  £(k,  m)  =  Z(n,  b). 


THE  CIRCLE. 


55 


132.  Definition.  An  angle  whose  vertex  is  on  a 
circle  of  which  its  sides  contain  chords  is  called  an 
inscribed  angle,  and  said  to  be  upon  the  chord  be- 
tween its  sides. 

133.  Theorem.  Inscribed  angles  upon  the  same 
chord  and  the  same  side  of  it  are  congruent. 

Proof.  1  st.  If  the  chord  BC  be  a  diameter  the 
straight  through  the  vertex  A  of 
any  inscribed  angle  and  the  cen- 
ter 0  makes  two  isosceles  tri- 
angles. .'.  bisecting  £BOF  we  b| 
get  ZDOFmtBAO.  In  same 
way  ZHOF^ZCAO.  :.  (by 
49)  *BAC  =  *DOH9  which  (by 
130)  is  right. 

2d.  If  the  vertex  A  be  on  the  same  side  of  the 
straight  BC  as  the  center  0, 
then  sect  OA  cannot  cut  BC, 
and  (by  III  1)  the  center  0 
is  between  A  and  the  other 
point  A'  of  the  circle  on  the 
straight  AO.  If  now  ray  OA 
be  costraight  with  a  side  of 
ifBOC,  then  aAOC  being  isos- 


Fig. 


50. 


Fig.  51. 


celes,  the  bisector  OF  of  ^  BOC 
makes  $F0C  =  ZBAC. 

Again  if  ray  OA'  is  within 
■4  BOC,  then  the  bisectors  OF 
and  OH  make 

lA'OF^ifOAB  and 

4A'0n=i.0AC.        ;.  (by  49) 
%-BAC  =^F0Ht  which 


Fig.  52. 


56  RATIONAL    GEOMETRY. 

(by  129)  is  congruent  to  each  angle  made  in  bisect 
ing  BOC. 
If,  however,  ray  OA'  is  without  ?{BOC,  then  the 
bisector  OD  of  4-  A'OB  makes 
^AVD=^OAB,   and   the  bi- 
sector   OE    of    ^-A'OC   makes 
tA'OEs  i-OAC.       .'.    (by  49) 
^BAC  =  4D0E,  which  (by  131) 
is  congruent  to  each  angle  made 
Fig.  53.  in  bisecting  BOC. 

3d.  If  the  vertex  A  and  the  center  0  be  on 
opposite  sides  of  the  straight  BC.  Let  Af  be  the 
other  point  of  the  circle  on  the  straight  AO.  Then 
the  six  angles  of  the  two  triangles  ABC,  A'BC  to- 
gether form  four  right  angles.  But  by  case  1st, 
the  two  angles  at  C  form  a  right  angle,  likewise 
the  two  at  B.  .'.  ^BAC  is  the 
supplement  of  ~4-BA'C. 

134.  Corollary  to  133.  The 
inscribed  angle  upon  a  diameter 
is  right. 

135.  Definition.      A  polygon 
whose  sides  are  congruent   and  Fig.  54? 
whose  angles  are  congruent  is  called  regular. 

136.  Definition.  A  polygon  whose  vertices  are 
concyclic  is  called  cyclic. 

137.  Corollary  to  133.  In  a  cyclic  quadrilateral 
the  opposite  angles  are  supplemental. 

138.  Theorem.  If  a  straight  have  one  point  in 
common  with  a  circle  and  be  not  perpendicular  to  the 
radius  to  that  point,  it  has  also  a  second  point  in  com- 
mon with  the  circle. 


THE  CIRCLE. 


51 


Let  the   straight   a  have    the    point  P  in  com- 
mon with  the  OC[CP]  and 
be  not  _L  to  CP. 

From  C  drop  CM  _L  to  a. 
From  M  on  a  set  off  MP' 
mMP. 

.-.CP'mCP.    .'.P'ison 
OC[CP].  Fig.  55. 

139.  Definition.  A  straight  which  has  two 
points  in  common  with  a  circle  is  called  a 
secant. 

140.  Theorem.  A  straight  perpendicular  to  a  diam- 
eter at  an  end-point  has  only  this  end-point  in  common 
with  the  circle. 

Proof.  Any  chord  is  (by  117)  bisected  by  the 
perpendicular  from  the  center. 

141.  Definition.  A  straight  which  has  only  one 
point  in  common  with  a  circle  is  called  a  tangent  to 
the  circle,  and  the  point  is  called  the  point  of 
contact. 

142.  Theorem.     If  BC  be  perpendicular  to  ABf 

iC  and   D  any  point   on   the 

straight  A  B  other  than  B, 
and  on  ray  CD  we  take  sect 
CF  =  CB,  then  F  is  within 

sect  CD. 

Proof.  Otherwise  since 
A  CBF  is  isosceles,  two  an- 
gles of  a  triangle  would  each  be  right  or  each  obtuse, 
which  (by  79)  is  impossible. 

143.  Theorem.  If  the  rays  of  one  angle  are  within 
another  the  angles  are  not  congruent. 


Fig.  56. 


58 


RATIONAL   GEOMETRY. 


Fig.  57. 
144.  Theorem. 


Proof.  For  suppose  £  (h,  m)  m  %  (k,  I)  and 
k,  I  within  4  (h,  m).  On 
the  other  side  of  m  from 
the  points  of  h  there  is  a 
ray  n  such  that  ~4-  (m,  n)~ 
7f(ht  k).  ;.  (by  49)  *(*,») 
=  ^.(h,  m).  .'.  from  our 
hypothesis  ^  (k,  n)  =  ^  (k,  l)7 
which  (by  III  4)  is  impos- 
sible. 
If  P  be  a  point  within  the  triangle 
ABC, then  angle  APC  is  not  congruent  to  angle  ABC. 

Proof.  The  ray  BP 
must  (by  30}  have  on  it  a 
point  D  within  the  sect 
AC.  :.PD  is  within 
ifAPC.  From  P  take 
PF  ||  BC.  It  makes 
4FPD  =  ^CBD.  From 
P  take  P^H^E.  It  makes 
.-.  (by  49)   ^-FPG  =  i-CBA. 

If  now  we  supposed  4  CBA  =  ^  CPA  we  should 
have  ^FPG  =  ifCPA,  which  (by  143)  is  impossible. 
145.  Theorem.  //  two  triangles  have  a  side  in  com- 
mon and  the  angles  opposite  it 
congruent,  and  with  vertices  on 
the  same  side  of  it,  the  four  ver- 
tices are  coney  die. 

Proof.     If  the  circle  through 

A,  B,  C  did  not  contain  D,  then 

(by  138)  it  would  have  a  second 

Fig.  59.  point     in     common    with    the 


D  A 

Fig.  58. 
■4GPD=^ABD. 


THE   CIRCLE. 


59 


straight  AD  or  CD,  or  else  we  would  have  OCljCD 
and  OALAD. 

But  in  this  last  case  the  whole  circle  except 
points  A  and  C  would  be  within  ^ADC.  For 
the  center  O  would  be 
within  ifADC,  being  then 
on  the  bisector  of  if  ADC 
since,  a  ADC  being  isos- 
celes, aAOD=aCOD  [3 
sides  =  ].  Hence  the  point 
B  would  be  within  a  ADC, 
which  (by  144)  is  impossi- 
ble. But  it  is  just  as  im- 
possible that  AD  or  CD 
(besides  A  or  C)  should  have  a  point 
the   circle   other  than    D.       For   then    We 


Fig.  60. 


on 


U 

would 
impos- 


have   ^.ADC^^AD'C,   which  (by   79)    is 
sible. 

146.  Theorem.  //  two  opposite  angles  of  a  quad- 
rilateral are  supplemental  it  is  cyclic. 

Proof.  Given  the  if  CD  A  is 
the  supplement  of  ^  B.  On  the 
circle  determined  by  A ,  B,  C  take 
a  point  D'  on  the  same  side  of 
AC  as  D.  Then  ^D'^^D, 
being  each  the  supplement  of 
ifB.  .'.  (by  145)  D  is  concylic 
with  A  CD\  that  is,  with 
ABC. 

147.  Corollary  to  146.  A  quadrilateral  is  cyclic 
if  an  angle  is  congruent  to  the  angle  adjacent  to 
its  opposite. 


Fig.  61. 


60  RATIONAL   GEOMETRY. 

Ex.  107.  Defining  a  tanchord  angle  as  one  between  a 
tangent  to  a  circle  and  a  chord  from  the  point  of  contact, 
prove  it  congruent  to  an  inscribed  angle  on  this  chord. 

Ex.  108.  An  angle  made  by  two  chords  is  how  related 
to  the  angles  at  the  center  on  chords  joining  the  end- 
points  of  the  given  chords? 

•  Ex.  109.  The    vertices    of    all  right-angled  triangles  on 
the  same  hypothenuse  are  coney clic. 

Ex.  no.  Tangents  to  a  circle  from  the  same  external 
point  are  congruent,  and  make  congruent  angles  with 
the  straight  through  that  point  and  the  center. 

Ex.  in.  Two  congruent  coinitial  chords  are  symmetric 
with  respect  to  the  coinitial  diameter. 

Ex.  112.  If  triangles  on  the  same  base  and  on  the  same 
side  of  it  have  the  angles  opposite  it  equal,  the  bisectors 
of  these  angles  are  copunctal. 

Ex.  113.  The  end-points  of  two  congruent  chords  of  a 
circle  are  the  vertices  of  a  symmetrical  trapezoid. 

Ex.  114.  The  chord  which  joins  the  points  of  contact 
of  parallel  tangents  to  a  circle  is  a  diameter. 

Ex.  115.  A  parallelogram  inscribed  in  a  circle  must  have 
diameters  for  diagonals. 

Ex.  116.  Of  the  vertices  of  a  triangle  and  its  ortho- 
center,  each  is  the  orthocenter  of  the  other  three. 

Ex.  117.  At  every  point  on  the  circle  can  be  taken  one, 
and  only  one,  tangent,  namely,  the  perpendicular  to  the 
radius  at  the  point. 

Ex.  118.  The  perpendicular  to  a  tangent  from  the  center 
of  the  circle  cuts  it  in  the  point  of  contact. 

Ex.  119.  The  perpendicular  to  a  tangent  at  the  point 
of  contact  contains  the  center. 

Ex.  120.  The  radius  to  the  point  of  contact  is  perpen- 
dicular to  the  tangent. 

Ex.  121.  An  inscribed  ||g'm  is  a  rectangle. 

Ex.  122.  The  bisector  of  any  ^  of  an  inscribed  quad' 
intersects  the  bisector  of  the  opposite  exterior  if  on  the  0. 

Ex.  123.  The  O  with  one  of  the  m  sides  of  a  -I-  A  as 
diameter  bisects  the  base. 


THE  CIRCLE.  61 

Ex.  124.  The  radius  is  =  to  the  side  of  a  regular  in- 
scribed  hexagon. 

Ex.  125.  If  the  opposite  sides  of  an  inscribed  quad'  be 
produced  to  meet,  the  bisectors  of  the  ^s  so  formed 
are    ±. 

Ex.  126.  The  circles  on  2  sides  of  a  A  as  diameters  in- 
tersect on  the  third  side  (in  the  foot  of  its  altitude). 

Ex.  127.  The  altitudes  of  a  A  are  the  ^  bisectors  of  its 
pedal  A   (the  feet  of  the  altitudes). 

Ex.  128.  AB  a  diameter;  AC  any  chord;  CD  tangent; 
BD±CD,  meets  AC  on  QB(BA). 

Ex.  129.  Find  a  point  from  which  the  three  rays  to  three 
given  points  make  =  ^  s. 

Ex.  130.  The  circumOs  of  3  As  made  by  3  points  on 
the  sides  of  a  A,  2  with  their  vertex,  are  copunctal. 


V.  The  Archimedes  Assumption.* 

V.  Let  Ax  be  any  point  on  a  straight  between  any 
given  points  A  and  B\  take  then  the  points  A2,  A3> 
A 4,  .  .  .  ,  such  that  Ax  lies  between  A  and  A2>  further- 
more A2  between  Ax  and  A3,  further  A3  between  A2 
and  A 4,  and  so  on,  and  also  such  that  the  sects  AAlf 
AiA2,  A2A31  A3A4,  .  .  .  ,  are  congruent;  then  in  the 
series  of  points  A2t  A3,  AAt  .  .  .  ,  there  is  always  such 
a  point  A  n,  that  B  lies  between  A  and  A  n. 

148.  This  postulate  makes  possible  the  introduc- 
tion of  the  continuity  idea  into  geometry.  We  have 
not  used  it,  and  will  not,  since  the  whole  of  the  ordi- 
nary school-geometry  can  be  constructed  with  only 
Assumptions  I-IV. 

*  Archimedis  Opera,  rec.  Heiberg,  vol.  I,  1880,  p.  11. 


CHAPTER  VI. 

PROBLEMS  OF  CONSTRUCTION. 

Existence  theorems  on  the  basis  of  assumptions  I-V, 
and  the  visual  representation  of  such  theorems  by 
graphic  constructions. 

Graphic  solutions  of  the  geometric  problems  by 
means  of  ruler  and  sect-carrier. 

149.  Convention.  What  are  called  problems  of 
construction  have  a  double  import.  Theoretically 
they  are  really  theorems  declaring  that  the  exist- 
ence of  certain  points,  sects,  straights,  angles,  circles, 
etc.,  follows  logically  by  rigorous  deduction  from  the 
existences  postulated  in  our  assumptions.  Thus  the 
possibility  of  solving  such  problems  by  elementary 
geometry  is  a  matter  absolutely  essential  in  the 
logical  sequence  of  our  theorems. 

So,  for  example,  we  have  shown  (in  101)  that  a 
sect  has  always  trisection  points,  and  this  may  be 
expressed  by  saying  we  have  solved  the  problem  to 
trisect  a  sect.  Now  it  happens  that  a  solution  of 
the  problem  to  trisect  any  angle  is  impossible  with 
only  our  assumptions.  Thus  any  reference  to  re- 
sults following  from  the  trisection  of  the  angle  would 
be   equivalent    to    the   introduction    of   additional 

assumptions. 

62 


PROBLEMS  OF  CONSTRUCTION.  63 

But  problems  of  construction,  on  the  other  hand, 
may  have  a  reference  to  practical  operations, 
usually  for  drawing  on  a  plane  a  picture  which 
shall  serve  as  an  approximate  graphic  repre- 
sentation of  the  data  and  results  of  the  existential 
theorem. 

Our  Assumptions  I  postulate  the  existence  of  a 
straight  as  the  result  of  the  existence  of  two  points. 
This  may  be  taken  as  authorizing  the  graphic  desig- 
nation of  given  points  and  the  graphic  operation 
to  join  two  designated  points  by  a  straight,  and  as 
guaranteeing  that  this  operation  can  always  be 
effected.  Confining  ourselves  to  plane  geometry,  on 
the  basis  of  the  same  Assumptions  I,  we  authorize 
the  graphic  operation  to  find  the  intersection-point 
of  two  coplanar  non-parallel  straights,  and  guaran- 
tee that  this  may  always  be  accomplished. 

To  practically  perform  these  graphic  operations, 
that  is  for  the  actual  drawing  of  pictures  which  shall 
represent  straights  with  their  intersections,  we  grant 
the  use  of  a  physical  instrument  whose  edge  is  by 
hypothesis  straight,  namely,  the  straight-edge  or 
ruler. 

Thus  Assumptions  I  give  us  as  assumed  con- 
structions, or  as  solved,  the  fundamental  problem 
of  plane  geometry: 

Problem  1.  (a)  To  designate  a  given  point  of  the 
plane; (b)  to  draw  the  straight  determined  by  two  points; 
to  -find  the  intersection  of  two  non-parallel  straights. 

150.  Our  Assumptions  III  postulate  the  existence 
on  any  given  straight  from  any  given  point  of  it  to- 
ward a  given  side,  of  a  sect  congruent  to  a  given  sect. 


64  RATIONAL   GEOMETRY. 

This  may  be  taken  as  authorizing  and  guarantee- 
ing the  graphic  operation  involved  in  what  may 
be  called 

Problem  2.  To  set  off  a  given  sect  on  a  given 
straight  from  a  given  point  toward  a  given  side. 

A  physical  instrument  for  actual  performance  of 
this  construction  in  drawing  might  be  called  a  sect- 
carrier.  Our  straight-edge  will  also  serve  as  sect- 
carrier  if  we  presume  that  the  given  sect  may  be 
marked  off  on  it,  and  it  then  made  to  coincide 
with  the  given  straight  with  one  of  the  marked 
points  in  coincidence  with  the  given  point  of  the 
straight. 

Notice  that  in  these  graphic  interpretations  we 
freely  use  the  terminology  of  motion,  while  the  real 
existential  theorems  themselves  are  independent  of 
motion,  underlie  motion,  and  explain  motion.  We 
assume  that  the  motion  of  our  physical  instruments 
is  rigid. 

151.  We  now  announce  the  important  theorem 
that  in  our  geometry  all  graphic  problems  can  be 
solved,  all  graphic  constructions  effected,  merely  by 
using  problems  1  and  2. 

Theorem.  Those  geometric  construction  problems 
{existential  theorems)  solvable  by  employing  exclu- 
sively Assumptions  I-V  are  necessarily  graphically 
solvable  by  means  of  ruler  and  sect-carrier. 

The  demonstration  will  consist  in  solving  with 
problems  1  and  2  the  three  following  problems : 

152.  Problem  3.  Through  a  given  point  to  draw  a 
parallel  to  a  given  straight. 

Given  the  straight  AB  and  the  point  P. 


PROBLEMS   OF  CONSTRUCTION. 


65 


Construction.  Join  P  with  any  point  A  of  AB  by 
Prob.  1.  On  the  straight 
PA  beyond  A  take  (by 
Prob.  2)  AC^AP.  JoinC 
with  any  other  point  B  of 
AB.  On  the  straight  CB 
beyond  B  take  BQ=BC. 
PQ  is  the  parallel  sought. 

Proof.     By  98. 

153.  Problem  4.  To  draw  a  perpendicular  to  a 
given  straight. 

Construction.     Let  A  be  any  point  of  the  given 

straight.  Set  off  from  A 
on  this  straight  toward 
both  sides  two  congruent 
sects,  AB  and  AC,  and 
then  determine  on  any 
two  other  straights 
through  A  the  points  E 
and  D,  on  the  same  side 
of  AB,  and  such  that 
AB  =  AD=AE.  Since  ^ABD  and  4 ACE  are 
angles  at  the  bases  of  isosceles  triangles,  .*.  they  are 
acute,  .'.  the  straights  BD  and  CE  meet  in  F,  and 
also  the  straights  BE  and  CD  in  H.  Then  FH  is 
the  perpendicular  sought. 

Proof.  -4-BDC  and  ^BEC,  as  inscribed  angles 
on  the  diameter  BC,  are  (by  134)  right.  Since  (by 
127)  the  altitudes  of  &BCF  are  copunctal,  .\  FH  is 
1  to  BC. 

154.  Problem  5.  To  set  off  a  given  angle  against 
a  given  straight,  or  to  construct  a  straight  cutting  a 
given  straight  under  a  given  angle. 


Fig.  63. 


66  RATIONAL   GEOMETRY. 

Given  /?  the  angle  to  be  set  off>  and  A  its 
vertex. 

Construction.  We  draw,  by  Prob.  3,  the  straight 
/  through  A  ||  to  the  given 
straight  against  which  the 
given  angle  fi  is  to  be 
set  off.  By  Prob.  4  draw  a 
straight  ±  to  /  and  a  straight 
J_  to  one  side  of  /?.  Through 
any  point  B  of  the  other  side 
of  ft  draw,  by  Prob.  3,  ||s  to 
these  J_s.  Call  their  feet  C  and  D.  Then  (by 
Probs.  4  and  3)  draw  from  A  a  st'  1  to  CD.  Call 
its  foot  E. 

Then  ^fCAE=t9.  So  EA  will  cut  the  given 
straight  ||  to  I  under  the  given  ^/?. 

Proof.  Since  ^-ACB  and  ifADB  are  right,  so 
(by  146)  the  four  points  A,  B,  C,  D  are  concyclic. 
Consequently  ^fACD=  ifABD  (by  133)  being  in- 
scribed angles  on  the  same^  chord  AD  and  on  the 
same  side  of  it.  Therefore  their  complements 
4CAE=  4  BAD. 

155.  This  completely  demonstrates  our  theorem, 
151,  since  the  existential  theorems  in  Assumptions 
II  guarantee  the  solution  of  problems  requiring  no 
new  graphic  operations,  such  as  to  find  a  point  within 
and  a  point  without  a  given  sect,  and  certain  other 
problems  of  arrangement;  while  Assumption  V 
would  simply  guarantee  the  finding  of  a  point  with- 
out a  given  sect  by  repeating  a  certain  specific  ap- 
plication of  our  Prob.  2. 

156.  In  our  geometry,   though  constantly  using 


PROBLEMS   OF  CONSTRUCTION.  67 

graphic  figures,  we  must  never  rely  or  depend  upon 
them  for  any  part  of  our  proof.  We  must  always 
take  care  that  the  operations  undertaken  on  a  figure 
also  retain  a  purely  logical  validity. 

157.  This  cannot  be  sufficiently  stressed.  In  the 
right  use  of  figures  lies  a  chief  difficulty  of  our  in- 
vestigation. 

The  graphic  figure  is  only  an  approximate  sug- 
gestive representation  of  the  data.  We  cannot  rely 
upon  what  we  suppose  to  be  our  immediate  per- 
ception of  the  relations  in  even  the  most  accurate 
obtainable  figure. 

In  rigorous  demonstration,  the  figure  can  be  only 
a  symbol  of  the  conceptual  content  covered  by  its 
underlying  assumptions. 

The  logical  coherence  should  not  be  dependent 
upon  anything  supposed  to  be  gotten  merely  from 
perception  of  the  figure.  No  statement  or  step  can 
rest  simply  on  what  appears  to  be  so  in  a  figure. 
Every  statement  or  step  must  be  based  upon  an 
assumption,  a  definition,  a  convention,  or  a  preced- 
ing theorem. 

Yet  the  aid  from  figures,  from  sensuous  intuition, 
is  so  inexpressibly  precious,  that  any  attempt  even 
to  minimize  it  would  be  a  mistake.-  That  treat- 
ment of  a  proposition  is  best  which  connects  it  most 
closely  with  a  visualization  of  the  figure,  while  yet 
not  using,  as  if  given  by  the  figure,  concepts  not 
contained  in  the  postulates  and  preceding  propo- 
sitions. 

158.  As  an  immediate  result  of  Prob.  5,  the  proofs 
in  Chapters  I-V  of  our  existential  theorems  give  ruler 


68 


RATIONAL    GEOMETRY. 


Fig.  65. 


and  sect-carrier  solutions  of  the  corresponding  prob- 
lems.    We  will  now  give  some  alternative  solutions. 

159.  Problem  6.  At  a  given  point  A  to  make  a 
right  angle. 

Solution.  Draw  through  A 
any  straight  AD,  and  through 
D  any  other  straight  BC,  and 
make  AD  =  BD  =  CD.  Then 
is  ifBAC  right.  [Inscribed 
angle  on  a  diameter.] 

160.  Problem  7.  From  a  given  point  A  to  drop  a 
perpendicular  upon  a  given  straight  BC. 

Solution.  By  Prob.  6,  at 
A  construct  a  rt.  i-BAC. 
Make  BD=BA.  Draw  DE 
\\AC.  Make  BF=BE. 
Then  is  AF±  BC. 

Proof.  ^ABF=  aDBE. 
[2  sides  and  inc.  %■  =  .] 

161.  Problem  8.  At  any  point  A  on  a  straight  BC 
to  erect  the  perpendicular. 

Solution.  By  Prob.  7,  from  any  point  without  the 
straight  drop  to  it  a  perpendicular.  By  Prob.  3, 
draw  a  parallel  to  this  through  A . 

162.  Problem  9.   To  bisect  a  given  sect  AB. 
Construction.  Draw  through 

B  any  other  straight  BC.  Make 
onitBC=CD  =  DE.  Produce 
AE^EF.  Draw  FDG.  Then 
is  AG=GB. 

Proof.  D  is  the  centroid 
of  aABF. 


Fig.  66. 


PROBLEMS  OF  CONSTRUCTION. 


69 


163.  Problem  5.  At  a  given  point  in  a  given 
straight  to  make  an  angle  congruent  to  a  given  angle. 
Required,  against  the  given 
ray  A B  of  a,  and  toward  a 
given  side  of  a,  the  C-side,  to 
make  an  angle  =  ^  D  (given) . 

Construction.      To  one  side 
of  the  given  acute  angle  erect 
(by  161)  FH  _L  DF,  meeting 
the  other  side  at  H.      Take  AB=DF  and  BCLAB 
and  BC^FH.     .-.  (by  III  6)  t£BAC=  ^-FDH. 

164.  Problem  10.  To  bisect  a  given  angle. 

Construction.     On  one 

side  of  the  given  ~4-A 
take  any  two  points  B, 
C.  On  the  other  side 
take  ABf  =  AB,  and  AC 
69-  =  AC.     The    sects    BC 

and  B'C  intersect,  say  at  D.     AD  is  the  desired  bi- 
sector. 

165.  Problem.    To  join  two  points  by  an  arc  con- 
taining a  given  angle. 

Let  A,  B  be  the  two  points,  a  the  given  angle. 
Make  an  angle  BAC  supplemental 
to  a.  Erect  the  perpendicular  to 
AC  at  A,  and  to  AB  at  the  bi- 
section-point. Their  point  of  in- 
tersection is  the  center  of  the 
required  circle.      ^  AFB  =  a. 

Proof.    Their    supplements 


Fig. 


tAOD^lBAC  (complements  of  ^OAD). 


7©  RATIONAL   GEOMETRY. 

1 66.  Problem.  To  describe  a  circle  touching  three 
given  intersecting  but  not  copunctal  straights. 

Construction.  At  the 
points  of  intersection  draw 
the  angle-bisectors.  From 
the  cross  of  any  two  of 
these  bisectors,  the  perpen- 
dicular upon  either  of  the 
three  straights  is  the  radius 
of  a  circle  touching  all 
three. 

167.  Definition.  The  cointersection-point  of  the 
three  bisectors  of  the  internal  angles  of  a  triangle,  /, 
is  called  the  triangle's  in-center  [r,  the  in-radius]; 
Ol(r)  the  in-circle. 

168.  Definition.  A  circle  touching  one  side  of  a 
triangle  and  the  other  two  sides  produced  is  called 
an  escribed  circle,  or  ex-Q.  The  three  centers  Il9 
I2y  I3  of  the  escribed  circles  Oli(rg),  Ol2{r2)t  OJ3(r3) 
of  a  triangle  are  called  its  ex-centers. 

Ex.  131.  A  right  angle  can  be  trisected. 

Ex.  132.  To  construct  a  triangle,  given  two  sides  and 
the  included  angle. 

Ex.  133.  To  construct  a  triangle,  given  two  angles  and 
the  included  side. 

Ex.  134.  To  construct  a  triangle,  given  two  angles  and 
a  side  opposite  one  of  them. 

Ex.  135.  To  describe  a  parallelogram,  given  two  sides 
and  the  included  angle. 

Ex.  136.  To  construct  an  isosceles  triangle,  having  given 
the  base  and  the  angle  at  the  vertex. 

Ex.  137.  To  erect  a  perpendicular  to  a  sect  at  its  end- 
point,  without  producing  the  sect  or  using  parallels. 

Hint.     At  this  end-point  against  the  given  sect  make 


PROBLEMS  OF  CONSTRUCTION.  71 

any  acute  angle.  At  any  other  point  of  the  sect  make 
toward  this  a  congruent  angle. 

Beyond  the  intersection-point  of  the  rays,  make  on 
this  second  ray  a  sect  congruent  to  a  side  of  this  isosceles 
triangle.     Its  end  is  a  point  of  the  required  perpendicular. 

Ex.  138.  Construct  a  circle  containing  two  given  points 
with  center  on  a  given  straight. 

Ex.  139.  To  draw  an  angle-bisector  without  using  the 
vertex. 

Ex.  140.  Through  a  given  point  to  draw  a  straight 
which  shall  make  congruent  angles  with  two  given 
straights. 

Ex.  141.  In  a  straight  find  a  point  with  which  two 
given  points  give  equal  sects. 

Ex.  142.  From  two  given  points  on  the  same  side  of  a 
straight  to  draw  two  straights  intersecting  on  it  and 
making  congruent  angles  with  it. 

Ex.  143.  To  draw  a  straight  through  a  given  point 
between  two  given  straights  such  that  they  intercept 
on  it  a  sect  bisected  by  the  given  point. 

Ex.  144.  Through  a  given  point  to  draw  a  st'  making 
=  ^s  with  the  sides  of  a  given  •£. 

Ex.  145.  Construct  +  A  from  b  and  hb',  from  a  and  b; 
from  /?  and  a -\-b;  from  fi  and  hb;  from  b  and  /?;  from  p 
and  hb]   from  p  and  a;   from  6  and  r. 

Ex.  146.  Construct  r't  A  from  a  and  hc;  from  a  and  c; 
from  c  and  r;  from  a  and  r;  from  /?  and  r;  from  a  and 
a  +  b;   from  R  and  r. 

Ex.  147.  Construct  A  from  p,  a,  and  ha;  from  p,  a, 
and  /?;  from  its  pedal;  from  b,  a  +  c,  a;  from  a,  hb,  p\ 
from  /„  I2t  I3. 

Ex.  148.  Without  prolonging  two  sects,  to  find  the 
bisector  of  the  ^  they  would  make. 

Ex.  149.  Describe  O  through  two  given  points  with 
center  on  given  st' ;   with  given  radius. 

Ex.  150.  From  one  end  of  the  hypothenuse  lay  off  a 
sect  on  it  congruent  to  the  ±  from  the  end  of  this  sect  to 
the  other  side. 


72  RATIONAL    GEOMETRY. 

Ex.  151.  From   -I-  A  cut  a  trapezoid  with  3  sides  =. 

Ex.  152.  To  inscribe  a  sq.  in  a  given  r't-l-  A. 

Ex.  153.  Find  point  in  side  of  -I-  A  where  J-  erected 
and  produced  to  other  side  is   =  to  base. 

Ex.  154.  To  describe  a  O  which  shall  pass  through  a 
given  point  and  touch  a  given  st'  at  a  given  point. 

Ex.  155.  AB,  AC,  BD,  CE,  are  chords.  BD  ||  AC, 
CE  ||  AB.     Then  AF  \\  DE  is  a  tangent. 

Ex.  156.  To  describe  a  O  whose  center  shall  be  in  one  ± 
side  of  a  r't  A  while  the  O  goes  through  the  vertex  of  the 
r't  ^  and  touches  the  hypothenuse. 

Ex.  157.  To  describe  a  O  of  given  radius  with  center 
in  one  side  of  a  given  ^  and  tangent  to  the  other  side. 

Ex.  158.  Construct  A  from  a,  a,  and  that  to.  trisects  a; 
from  a  and  orthocenter;   from  a  and  centroid. 

Ex.  159.  Construct  A  from  a,  /?,  R;  from  feet  of  medians; 
of  altitudes. 

Ex.  160.  (Brahmagupta.)  If  the  diagonals  of  an  in- 
scribed quad'  are  ±,  the  st'  through  their  intersection 
_L  to  any  side  bisects  the  opposite  side. 


CHAPTER  VII. 

SIDES,  ANGLES,  AND  ARCS. 

169.  Convention.  When  a  sect  congruent  to  CD 
is  taken  on  sect  A  B  from  A  and  its  second  end- point 
falls  between  A  and  B,  then  A B  is  said  to  be  greater 
than  CD;  (AB>CD).  When  an  angle  congruent 
to  4  (hf  k\  is  set  off  from  vertex  0  against  one  of 
the  rays  of  ^.AOB  toward  the  other  ray,  if  its 
second  side  falls  within  4-AOB,  then  ^.AOB  is  said 
to  be  greater  than  ^  (h,  k).     In  symbols,  ?fAOB> 

170.  Theorem.  //  the  first  side  of  a  triangle  be 
greater  than  a  second,  then  the  angle  opposite  the  first 
must  be  greater  than  the  angle  opposite  the  second. 

Given  BA>BC. 
To  prove  ^C>  ^A. 
Proof.    From  B  toward  A  take 
BD  =  BC.     The  end- point  D  of 

r IG.    72. 

this  sect  then,  because  BA  >  BC, 
is  between  A  and  B,  that  is  within  ^-ACB,  as  is 
therefore    also    CD.       Then    is     aBDC    isosceles, 
.-.  tCDB=  i-DCB.    ;.  ^ACB>  4 BCD  or  ^.CDB. 
But  (by  79)  4-CDB>  4 A. 

73 


74 


RATIONAL   GEOMETRY, 


171.  Inverse.     If  ^A>  7{B,  .'.a>b. 
Proof.     [From  57  and  170.] 

172.  Definition.  Except  the  perpendicular,  any 
sect  from  a  point  to  a  straight  is  called  an  oblique. 

173.  Theorem.  From  a  point  to  a 
straight  any  oblique  is  greater  than  the 
perpendicular. 

Proof.  Since  ^-CAB  is  r't,  .'.  (by  79) 
Fig-  73.        -4A>  ^B.     .*.  (by  171)  a>b. 

174.  Theorem.     Any  two  sides  of  a  triangle  are 
together    greater     than     the 
third  side. 

Proof.  On  st'  BC,  be- 
yond C,  take  CD^CA. 
•'•  (by  57)  i-CDA^^CAD. 
But  AC  is  within  *  DAB, 
:.  ^-DAB>  ^DAC=ifD. 
.'.  (by  171)  BD>AB. 

175.  Theorem.  (The  ambiguous  case.)  //  two  tri- 
angles have  two  sides  of  the  one  congruent  respectively 
to  two  sides  of  the  other,  and  the  angles  opposite  one 
pair  of  congruent  sides  congruent,  then  the  angles  oppo- 
site the  other  pair  are  either  congruent  or  supplemental. 


Fig.  74. 


c  A  H 

Fig.  75. 


Hypothesis,     a  ABC  and  a  FGH  with  ^  A  =  4  F, 
AB^FG,  and  BC^GH. 


SIDES,  ANGLES,  AND   ARCS.  75 

Conclusion.  ~4-C=  ?fH}  or  ^.C  supplement  of 
*H. 

Proof.  At  B  against  BA  take,  on  the  side  toward 
C,  the  ^ABC'  =  ^G.  If  ray  BC  falls  on  ray  BC, 
then  (by  80)  ^fC=  %-H.  If  not  on  BC,  suppose  C 
between  C  and  A.  Then  (by  44)  ^fBCA^^fH, 
and  BC'mGHmBC.     .'.  (by  57)  ^BCC=  *C. 

176.  Corollary  to  175.  Two  triangles  are  con- 
gruent if  they  have  two  sides  and  the  angle  opposite 
the  greater  respectively  congruent. 

177.  Definition.  A  triangle  one  of  whose  angles 
is  a  right  angle  is  called  a  right-angled  triangle, 
or  more  briefly  a  right  triangle.  The  side  opposite 
the  right  angle  is  called  the  hypothcnuse. 

178.  Corollary  to  176.  Two  right-angled  trian- 
gles are  congruent  if  the  hypothenuse  and  one  side 
are  respectively  congruent. 

Ex.  161.  If  two  triangles  have  two  sides  of  the  one  respec- 
tively congruent  to  two  sides  of  the  other,  and  the  angles 
opposite  one  pair  of  congruent  sides  congruent,  then  if 
these  angles  be  not  acute  the  triangles  are  congruent. 

Ex.  162.  If  two  triangles  have  two  sides  of  the  one 
respectively  congruent  to  two  sides  of  the  other,  and 
the  angles  opposite  one  pair  of  congruent  sides  congruent, 
then  if  one  of  the  angles  opposite  the  other  pair  of  con- 
gruent sides  is  a  right  angle  the  triangles  are  congruent. 

Ex.  163.  If  two  triangles  have  two  sides  of  the  one 
respectively  congruent  to  two  sides  of  the  other,  and 
the  angles  opposite  one  pair  of  congruent  sides  congruent, 
then  if  the  side  opposite  the  given  angle  is  congruent  to 
or  greater  than  the  other  given  side  the  triangles  are 
congruent. 

Ex.  164.  If  any  triangle  has  one  of  the  following  proper- 
ties it  has  all: 


76  RATIONAL   GEOMETRY. 

i.  Symmetry. 

2.  Two  congruent  sides. 

3.  Two  congruent  angles. 

4.  A  median  which  is  an  altitude. 

5.  A  median  which  is  an  angle-bisector. 

6.  An  altitude  which  is  an  angle-bisector. 

7.  A  perpendicular  side-bisector  which  contains  a 
vertex . 

8.  Two  congruent  angle-bisectors. 

Ex.  165.  The  difference  of  any  two  sides  of  a  triangle 
is  less  than  the  third  side. 

Ex.  166.  From  the  ends  of  a  side  of  a  triangle  the  two 
sects  to  a  point  within  the  triangle  are  together  less  than 
the  other  two  sides  of  the  triangle,  but  make  a  greater 
angle. 

Ex.  167.  Two  obliques  from  a  point  making  congruent 
sects  from  the  perpendicular  are  congruent,  and  make 
congruent  angles  with  the  straight. 

Ex.  168.  Of  any  two  obliques  between  a  given  point 
and  straight  that  which  makes  the  greater  sect  from  the 
foot  of  the  perpendicular  is  the  greater. 

Ex.  169.  Of  sects  joining  two  symmetrical  points  to  a 
third,  that  cutting  the  axis  is  the  greater. 

179.  Theorem.     //  two  triangles  have  two  sides  of 
the  one  respectively  congruent  to  two  sides  of  the  other, 
then  that  third  side  is  the  greater 
which    is    opposite    the    greater 
angle. 

Proof.  Take  the  triangles 
with  one  pair  of  congruent 
sides  in  common,  BC,  and  on 
the  same  side  of  BC  the  other 
pair  of  congruent  sides,  BA,  BA' '.  The  bisector 
of  i-ABA',  being  within  ^fABC,  meets  AC  at  a 
points.     Then  (by  43)   aABG  =  aA'BG.      .'.AG 


SIDES,  ANGLES,  AND  ARCS.  77 

=  j4'G\      But  (by   174)   A'G  and  GC  are  together 
greater  than  A'C . 

179''.  Inverse  of  179.  If  two  triangles  have  two 
sides  of  the  one  respectively  congruent  to  two  sides 
of  the  other,  then,  of  the  angles  opposite  their  third 
sides,  that  is  the  greater  which  is  opposite  the 
greater  third  side. 

Ex.  170.  Two  right  triangles  are  congruent  if  the  hy- 
pothenuse  and  an  acute  angle  are  congruent,  or  if  a  per- 
pendicular and  an  acute  angle  are  congruent  to  a  per- 
pendicular and  the  corresponding  acute  angle. 

Ex.  171.  Given  AB  a  sect,  C  its  bisection-point,  PA  m 
PB. 

Prove  PC  LAB. 

Ex.  172.   Inverse.   Given  CPJLbi'  of  AB.    Prove  PA=PB. 

Ex.173.  Given  PM±AM=PN±AN.  Prove^.PAM  = 
^  PAN  or  its  complement. 

Ex.  174.  Inverse.  Given  ifPAM  =  ^PAN.  Prove 
PM  ±  AM  mPNL  AN. 

180.  Definition.  If  AB  is  a  diameter  of  a  circle 
with  center  C,  then  the  two  points  of  the  circle  on 
any  other  diameter,  being  on  opposite  sides  of  C, 
are  (by  25)  on  opposite  sides  of  the  straight  AB. 
Hence  the  points  of  the  circle  other  than  the 
points  A}  B,  are  separated  by  AB  into  two 
classes  of  points  uniquely  paired.  One  of  these 
classes  together  with  the  point  A  is  called  a 
semicircle.  The  other,  with  B}  is  the  associated 
semicircle;  A  and  B  are  called  end-points  of  each 
semicircle. 

181.  Definition.  If  A,  D  are  two  points  on  the 
circle  with  center  C,  then,  since  (by  142)  the  end  of 


78  RATIONAL    GEOMETRY. 

the  perpendicular  from  C  to  the  straight  AD  falls 
within  a  radius,  therefore  the 
points  of  the  circle  are  not  all 
on  the  same  side  of  the  secant 
AD.  Hence  the  points  of  the 
circle  other  than  the  points 
A,  D  are  separated  by  AD  into 
two  classes  of  points.  One  of 
these  classes,  together  with 
the  point  A,  is  called  an  arc.  The  other,  with  the 
point  D,  is  called  the  associated  or  explemental  arc. 
A  and  D  are  called  end-points  of  each  arc. 

Of  these  two  arcs  the  arc  on  the  side  oi  AD  re- 
mote from  the  center  is  called  the  minor  arc.  The 
arc  on  the  same  side  of  AD  as  the  center  is  called 
the  major  arc.  The  chord  AD  is  said  to  be  the 
chord  of  each  of  the  two  arcs.  Thus  to  every  arc 
pertains  a  chord,  and  to  every  chord  pertain  two 
arcs. 


Fig.  78. 
182.  Definition.  Two  arcs  AB,  A'B',  are  called 
congruent  when,  the  end-points  being  mated,  to 
every  point  C  of  the  first  arc  corresponds  one,  and 
only  one,  point  C  of  the  second,  such  that  AC=A'C 
and  BC=B'C. 


SIDES,  ANGLES,  AND  ARCS.  79 

183.  Corollary  to  182.  Congruent  arcs  have  con- 
gruent chords. 

184.  Definition.  An  angle  having  its  vertex  at 
the  center  of  the  circle  is  called  an  angle  at  the  center, 
and  is  said  to  intercept  the  arc  and  chord,  whose 
end  points  are  on  the  angle's  sides  and  whose  other 
points  are  within  the  angle. 

185.  Theorem.  In  a  circle  or  in  circles  with  con- 
gruent radii,  congruent  angles  at  the  center  intercept 
congruent  arcs. 


Fig.  79. 

Given  tACBmtA'C'B'. 

To  prove  the  minor  arc  AB  =  minor  arc  A 'B'. 

Proof.  Since  (by  43)  *ACBm  aA'C'B';  :.AB  = 
A'B'. 

Moreover,  if  D  is  any  point  within  arc  AB  then 
ray  CD  is  within  ^  ACB.  Hence  (by  48)  there  is 
within  *  A'C'B'  a  ray  CD'  meeting  arc  A'B9  in  D' , 
which  makes  *  A'CUm  *  ACD  and  i£B'CD'  = 
*BCD.     :.  (by  43)  A'D'^AD  and  B'D'  =  BD. 

Also  any  point  D"  of  the  minor  arc  A'B'  such  that 
A'D"  =  AD  would  (by  58)  be  on  the  ray  making 
4A'CD"  =  4  ACD  =  tA'C'D',  and  hence  identical 
with  D'. 


8o 


RATIONAL    GEOMETRY. 


1 86.  Corollary 

187.  Theorem. 
gruent  radii. 

Given  arc  AMB  =  arc  A'M'B 
To  prove  CA=C'A'. 


Any  arc  may  be  bisected. 
Any  two  congruent  arcs  have  con- 


Fig.  80. 


Proof.  The  bisector  of  4  ACB  cuts  arc  AMB  in 
a  point  M  such  that,  (by  43)  aACM=aBCM. 
.'.  AM^BM  and  if  BMC=  4  AMC.  From  hypoth- 
esis there  is  a  point  M'  of  arc  A'M'B'  such  that 
a  A'M'B'  m  a  AMB.     .' .  A'M'  m  B'M'. 

.-.  (by  58)  aA'M'C'^aB'M'C. 

.'.  4A'M'C'^4B'M'C. 

.'.  (by  48  and  84)   ifA'M'C'  =  4  AMC. 

.'.  (by  44)  the  two  isosceles  triangles  aAMC  = 
aA'M'C.     .'.AC^A'C. 

188.  Inverse  of  185.  Congruent  minor  arcs  are 
intercepted  by  congruent  angles  at  the  center. 

Proof.    ,  Since  from  hypothesis  chord  AB  =  chord 
A'B'}  .-.  (by  187  and  58)  aACB^  aA'C'B'. 
.'.  ifACB^ifA'C'B'. 

189.  Theorem.  In  a  circle  or  in  circles  with  con- 
gruent radii,  congruent  chords  have  congruent  minor 
arcs. 


SIDES,  ANGLES,  AND  ARCS.  81 

For  the  angles  at  the  center  on  the  congruent 
chords  are  congruent  (by  58)  [As  with  3  sides  =  ]. 
.'.  (by  185)  the  minor  arcs  they  intercept  are  con- 
gruent. 

190.  Theorem.  Given  a  minor  arc  and  a  circle  of 
congruent  radius.  There  are  on  the  circle  two  and 
only  two  arcs  with  a  given  end-point,  congruent  to  the 
given  arc. 

Proof.  An  angle  at  the  center  which  intercepts 
the  given  arc  can  be  set  off  (by  III  4)  once  and 
only  once  on  each  side  of  the  radius  to  the  given 
point. 

191.  Theorem.  From  any  point  of  a  circle  there 
are  not  more  than  two  congruent  chords,  and  the  chords 
are  congruent  in  pairs,  one  on  each  side  of  the  diameter 
from  that  point. 

Proof.  If  AB  is  any  chord, 
take  at  center  C  on  the  other  side 
of  AC,  the  ^ACBf=^fACB, 
.-.by  43,  ^ACB'=  &ACB. 

:.ABfmAB. 

Moreover,  were  B"  the  end- 
point  of  a  third  chord  from  A 
congruent  to  AB  and  to  ABf , 
then  B,  B' ,  B"  would  be  at  once  on  OC(CA)  and 
OA(AB),  which,  by  122,  is  impossible. 

192.  Definition.  If  all  the  points  of  one  arc  are 
points  of  a  second,  but  the  second  has  also  points 
not  on  the  first,  then  the  second  is  said  to  be  greater 
than  the  first  and  any  arc  congruent  to  the  second 
is  said  to  be  greater  than  any  arc  congruent  to  the 
first. 


82  RATIONAL   GEOMETRY. 

193.  Theorem.  In  a  circle  or  in  circles  with  con- 
gruent radii,  of  two  angles  at  the  center,  the  greater 
intercepts  the  greater  arc  and  chord. 

Hypothesis.     CA^CA'.     4ACD>  4 A'CB'. 

Conclusion.     Arc  AD>  arc  A'B'. 


Fig.  82. 

Proof.  From  C  against  CA  toward  D,  (by  III  4) 
take  ifACB=  ^-A'C'B'.  Then  from  hypothesis 
ray  CB  is  within  ^fACD. 

.*.  B  is  within  arc  AD. 

.-.  (by  192)  arc  AD >  arc  AB.  But  (by  185) 
arc  AfB'  =  avc  AB.  .'.  (by  192)  arc  AD>  arc 
A'B'. 

Moreover  a  A'CB'  has  two  sides  CA',  CB'  =  CA, 
CD  of  aACD,  but  ^fACD>  *  A'CB',  ,\  (by  179) 
ADyA'B'. 

194.  Inverse  of  193.  In  a  circle  or  in  circles 
with  congruent  radii  the  greater  chord  has  the 
greater  angle  at  the  center  and  the  greater  minor 
arc. 

For  (by  i79&)  it  has  the  greater  angle  at  the  cen- 
ter, and  .*.  ,  by  193,  the  greater  minor  arc. 

195.  Inverse  of  193.  In  a  circle  or  in  circles 
with  congruent  radii,  the  greater   minor   arc   has 


SIDES,  ANGLES,  AND  ARCS.  83 

the  greater  angle   at   the   center   and   the  greater 
chord. 

196.  Theorem.  In  a  circle  or  in  circles  with  con- 
gruent radii,  congruent  chords  have  congruent  perpen- 
diculars from  the  center,  and  the  lesser  chord  has  the 
greater  perpendicular. 

Proof.  Of  two  chords  from  A  on  the  same  side  of 
the  diameter  AC,  one,  say  AD,  is  (by  III  4)  without 
the  angle  CAB  made  by  the 
other,  and  hence  its  end-point 
D  is  on  the  minor  arc  A  B.  Hence 
(by  195)  ^-ACB>  ^ACD  and 
AB>AD. 

Moreover,  the  sect  from  the 
center  to  the  bisection-point  of 
AD,  since  D  and  so  every  point 
of  AD  is  on  the  opposite  side  of  AB  from  C,  crosses 
the  straight  AB  and  .'.  (by  142)  is  >  the  perpen- 
dicular from  C  to  AB. 

Moreover,  congruent  chords  anywhere  have  con- 
gruent perpendiculars  (by  178). 

197.  Inverse  of  196.  In  a  circle  or  in  circles  with 
congruent  radii,  chords  having  congruent  perpen- 
diculars from  the  center  are  congruent,  and  the 
chord  with  the  greater  perpendicular  is  the  lesser. 
For  (by  196)  it  cannot  be  greater  nor  congruent. 


Two  Circles. 

198.  A  figure  formed  by  two  circles  is  symmetrical 
with  regard  to  their  center-straight  as  axis. 


84  RATIONAL   GEOMETRY. 

Every  chord  perpendicular  to  this  axis  is  bisected 
by  it. 

If  the  circles  have  a  common  point  on  this  straight 
they  cannot  have  any  other  point  in  common,  for 
any  point  in  each  has  in  that  its  symmetrical  point 
with  regard  to  this  axis,  and  circles  with  three  points 
in  common  are  identical. 

199.  Two  circles  with  one  and  only  one  point  in 
common  are  called  tangent,  are  said  to  touch,  and 
the  common  point  is  called  the  point  of  tangency 
or  contact. 

200.  If  two  circles  touch,  then,  since  there  is  only 
one  common  point,  this  point  of  contact  is  on  the 
center-straight,  and  a  perpendicular  to  the  center- 
straight  through  the  point  of  contact  is  a  common 
tangent  to  the  two  circles. 

Ex.  175.  Two  circles  cannot  mutually  bisect. 

Ex.  176.  The  chord  of  half  a  minor  arc  is  greater  than 
half  the  chord  of  the  arc. 

Ex.  177.  In  a  circle,  two  chords  which  are  not  both 
diameters  do  not  mutually  bisect  each  other. 

Ex.  178.  All  points  in  a  chord  are  within  the  circle. 

Ex.  179.  Through  a  given  point  within  a  circle  draw  the 
smallest  chord. 

Ex.  180.  Rays  from  center  to  intersection  points  of  a 
tangent  with  ||  tangents  are  JL. 

Ex.  181.  A  circle  on  one  side  of  a  triangle  as  diameter 
passes  through  the  feet  of  two  of  its  altitudes. 

Ex.  182.  In  +  a  ABC  HD  on  AB^BC,  prove  CD>AD. 

Ex.  183.  A  circumscribed  parallelogram  is  a  rhombus. 

Ex.  184.  In  &ABC,  having  AB>  BC,  the  median  BD 
makes  ^  BDA  obtuse. 

Ex.  185.  If  AB,  a  side  of  a  regular  A,  be  produced 
to  D,  then  AD>  CD>  BC. 

Ex.  186.   If  BD  is  bisector  fe,  and  AB>  BC,  then  BO  CD. 


SIDES,  ANGLES,  AND  ARCS.  85 

Ex.  187.  How  must  a  straight  through  one  of  the 
common  points  of  two  intersecting  circles  be  drawn  in 
order  tha>t  the  two  circles  may  intercept  congruent  chords 
on  it? 

Ex.  188.  Through  one  of  the  points  of  intersection  of 
two  circles  draw  the  straight  on  which  the  two  circles 
intercept  the  greatest  sect. 

Ex.  189.  If  any  two  straights  be  drawn  through  the 
point  of  contact  of  two  circles,  the  chords  joining  their 
second  intersections  with  each  circle  will  be  on  parallels. 

Ex.  190.  To  describe  a  O  which  shall  pass  through  a 
given  point,  and  touch  a  given  O  in  another  given 
point. 

Ex.  191.  To  describe  a  O  which  shall  touch  a  given 
O,  and  touch  a  given  st'  [or  another  given  O]  at  a  given 
point. 

Ex.  192.  The  foot  of  an  altitude  bisects  a  sect  from 
orthocenter  to  circum-O. 

Ex.  193.  If  from  the  end-points  of  any  diameter  of  a 
given  O  J_s  be  drawn  to  any  secant  their  feet  give  with 
the  center    ■   sects. 

Ex.  194.  A,  B,  I,  h  are  concylic. 

Ex.  195.  If  tb  meets  circum-O  in  D,  then  DA  =DC  =DI. 

Ex.  196.  The  _l_s  at  the  extremities  of  any  chord 
make   =  sects  on  any  diameter. 

Ex.  197.  If  in  any  2  given  tangent  Os  be  taken  any 
2  ||  diameters,  an  extremity  of  each  diameter,  and  the 
point  of  contact  shall  be  costraight. 

Ex.  198.  If  2  Os  touch  internally,  on  any  chord  of  one 
tangent  to  the  other  the  point  of  contact  makes  sects 
which  subtend  =  ^  s  at  the  point  of  tangency  of  the  Os. 

Ex.199.  2m0>  =  <a  according  as  if  A  acute,  r't, 
obtuse. 

Ex.  200.  Chords  joining  the  end-points  of  ||  chords 
are  ■„' 

Ex.  201.  St'  through  point  of  tangency  meets  O  0  at  A, 
OO'  at  A'.     Prove  AO  II  A'O'. 


86  RATIONAL   GEOMETRY, 

Ex.  202.  Intersecting  Os  are  •!•  with  regard  to  their 
center-st'  and  if  ■  are  \  with  regard  to  their  common 
chord. 

Ex.  203.  Find  the  hi*  of  an  ^  without  using  its  vertex. 

Ex.  204.  A  quad'  with  2  sides  ||  and  the  others  =  is 
either  a  ||g'm  or  a  symtra. 


CHAPTER  VIII. 


A   SECT    CA1CULUS. 


201.  On  the  basis  of  assumptions  I  i,  2, and  II-IV, 
that  is,  in  the  plane  and  without  the  Archimedes 
assumption,  we  will  establish  a  sect  calculus  or 
geometric  algorithm  for  sects,  where  all  the  oper- 
ations are  identical  with  those  for  real  numbers. 
The  following  proof  is  due  to  F.  Schur. 

202.  (Pascal.)  Let  A,B,C  and  A',  B',  C  be  two 
triplets  of  points  situated  respectively  on  two  per- 
pendiculars and  distinct  c' 
from  their  intersection 
point  0' .  If  AB'  is  par-  4 
allel  to  A  fB  and  BC 
parallel  to  B'C,  then  is 
also  AC  parallel  to  A'C. 

1  Proof.  Call  D'  the 
point  where  the  perpen- 
dicular from  B  upon  the 
straight  A'C  meets  the 
straight  B'A\  Then  C 
is  the  orthocenter  of  the  ,-// 
triangle  BA'D' ;  therefore  Jr 
D'CLA'B  and  .'.  ±ABf.  FlG-  84- 

Consequently  C  is  also  the  orthocenter  of  the  tri- 

87 


88  RATIONAL   GEOMETRY. 

angle  AB'D' \  :.  AD'lB'C  and  ,\  also  ±BC.  Con- 
sequently B  is  the  orthocenter  of  triangle  AC'D'\ 
\\  AC'±_D'B  and  .'.  AC'\\A'C. 

203.  Instead  of  the  word  "congruent"  and  the 
sign  = ,  we  use,  in  this  sect  calculus,  the  word 
"equal"  and  the  sign   =. 

204.  We  begin  by  showing  how  from  any  two 
sects  to  find  unequivocally  a  third  by  an  operation 
we  will  call  addition. 

205.  If  A ,  B,  C  are  three  costraight  points,  and  B 
lies  between  A  and  C,  then  we  designate  c=AC  as 
the  sum  of  the  two  sects  a=AB  and  b=BC,  and 
write  to  express  this  c  =  a  +  b. 


B 
j. g. »L ft 


* 


F c=a+b 

Fig.  85. 

To  add  the  two  sects  a  and  b  in  a  determined 
order,  we  start  from  any  point  A ,  and  take  the  point 
B  such  that  AB  =  t  that  is  =  a.  Then  on  the  straight 
AB  beyond  B  we  take  the  point  C  such  that  BC  =  b. 
Then  the  sect  AC  is  what  we  have  designated  as  the 
sum  of  the  two  sects  a=AB  and  b=BC  in  the  order 
a  +  b. 

206.  From  III  3  follows  immediately  that  this 
sum  is  independent  of  the  choice  of  the  point  A, 
and  independent  of  the  choice  of  the  straight  AB. 

By  III  1,  it  is  independent  of  the  order  in  which 
the  sects  are  added.     Therefore  a  +  b=b  +  a. 

207.  This  is  the  commutative  law  for  addition. 
Thus  the  commutative  law  for  addition  holds  good, 


A  SECT  CALCULUS.  89 

is  valid,  for  our  summation  of  sects.  But  this  law 
is  not  at  all  self-evident,  and  expresses  no  general 
magnitude  relation,  but  a  wholly  definite  geometric 
fact ;  for  a,  b  are  throughout  not  numbers,  but  only 
symbols  for  certain  geometric  entities,  for  sects. 

208.  The  sects  a  and  b  are  called  less  than  C;  in 
symbols :  a  <  c,  b  <  c ;  and  c  is  called  greater  than  a 
and  b;  in  symbols  c>a,  c>b. 

209.  To  add  to  a  +  b  a  further  sect  c,  take  on 
straight  A B  beyond  C  the  sect  CD=c.  Then  the 
sect  AD  =  (a  +  b)  +c.  But  this  same  sect  AD  is,  by 
the  given  definition  of  sum,  also  the  sum  of  the  sects 
AB  and  BD,  that  is  of  the  sects  a  and  (b  +  c). 

Thus  a  +  (b  +  c)  =  (a  +  6)  +  c,  and  so  is  verified  and 
valid  what  is  called  the  associative  law  for  addition. 

210.  To  define  geometrically  the  product  of  a  sect 
a  by  a  sect  6,  we  employ  the  following  construction. 
We  choose  first  an  arbitrary  sect,  which  remains  the 
same  for  this  whole  theory,  and  designate  it  by  1. 
This  we  set  off  from  their  inter- 
section point  on  one  of  two  per- 
pendicular straights.  On  the 
other  we  set  off  on  one  ray  a,  on 
the  other  b.  The  circle  through 
the  free  end-points  of  1,  a,  b  de- 
termines on  the  fourth  ray  a  sect 
c.     Then  we  name  this  sect  c  the 

product  of  the  sect  a  by  the  sect  b;  and  we  write 
c  =  ab. 

By  s  As  and  133,  ab=ba.  This  is  the  commuta- 
tive law  for  multiplication. 

211.  Considering  the  triangle  of  the  end-points  of 


9o 


RATIONAL    GEOMETRY. 


i  and  a,  it  is  equiangular  to  that  of  the  end- points 
of  b  and  c.  This  gives  as  an  easy  construction  for 
our  sect  product  the  following: 

Set  off  on  one  side  of  a  right  angle,  starting  from 
the  vertex  0,  first  the  sect  i  and  then,  likewise  from 
the  vertex  0,  the  sect  b.  Then 
set  off  on  the  other  side  the 
sect  a.  Join  the  end- points 
of  the  sects  i  and  a,  and  draw 
a  parallel  to  this  straight 
through  the  end-point  of  the 
sect  b.  The  sect  which  this 
FlG*  87  parallel    determines    on    the 

other  side  is  the  product  ab ;  or  we  may  call  it  ba, 
since,  as  we  have  already  seen,  ab=ba,  which  is 
also  given  by  the  fact  that  the  triangle  of  the  end- 
points  of  i  and  b  is  equiangular  to  that  of  the 
end-points  of  a  and  c. 

212.  We  emphasize  that  this  definition  is  purely 
geometric;  ab  is  not  at  all  the  product  of  two  num- 
bers. 

213.  To   prove  for  our  multiplication  the  asso- 


da=(bc)  a 


A  SECT  CALCULUS. 


91 


dative  law  for  multiplication  a  (be)  =  (ab)c  we  con- 
struct first  the  sect  d  =  bc}  then  da,  further  the 
sect  e=ba,  and  finally  ec.  The  end-points  of  da 
and  ec  coincide  (by  Pascal),  and  by  the  commuta- 
tive law  follows  the  above  formula  for  the  associa- 
tive law  of  sect  multiplication. 

214.  Finally  is  valid  in  our  sect-calculus  also 
the  distributive  law  a(b  +  c)  =ab  +  ac. 

To  demonstrate  it  we  construct  the  sects  ab,  ac, 
and  a(b  +  c),  and  draw  through  the  end-point  of 
the  sect  c  (see  Fig.  89)  a  parallel  to  the  other  side 
of  the  right  angle.  The  congruence  of  the  two 
right-angled  triangles  shaded  in  the  figure  and 
the  application  of  the  theorem  of  the  equality  of 


Xb+c) 


b+c 


opposite  sides  in  the  parallelogram  give  then  the 
desired  proof. 

215.  If  b  and  c  are  any  two  sects,  there  is  always 
one  and  only  one  sect  a  such  that  c  =  ab;  this  sect 

c 
a  is  designated  by  the  notation  j-t  and  is  called  the 

quotient  of  c  by  b. 


9 2  RATIONAL    GEOMETRY. 

The  Sum  of  Arcs. 

216.  Definition.  If  a  and  b  are  two  arcs  with 
equal  radii,  their  sum,  a  +  b,  is  the  arc  obtained 
by  taking  together  as  one  arc  the  arc  a  and  an  arc 
congruent  to  b  having  as  one  of  its  end- points  an 
end- point  of  a  and  its  points  taken  as  outside  of  a. 

217.  Theorem.  In  the  same  circle  or  in  circles 
with  equal  radii,  if  minor  arc  a  m  minor  arc  a!  and 


Fig. 


90. 


minor  arc  6  =  minor  arc  b' ,  then  arc  (a +  6)= arc 
(a' +  6'). 

Let  minor  arc  AB  =  a  and  minor  arc  BD=bf 
minor  arc  A'B'  =a!  and  minor  arc  B'D' =br. 

To  prove  arc  ABD= arc  A'B'D'. 

Proof.  '*CBF=&C'B'F'  (two  sides  and  in- 
cluded 4)  .\*CBF=tC'B'F'.  In  same  way 
4CBH^  ^C'B'H'.  .'.  (by  49)  ^HBF=  ^H'B'F'. 
;.  (two  sides  and  included  ^)  chord  A  D=  chord 
A'U .  .'.  (by  189)  minor  arc  AD  =  minor  arc  A'D'\ 
and  if  a  -j-  b  is  minor,  so  (by  inverse  of  133)  is  a'  +  6'. 
But  if  a  +  b  be  not  a  minor  arc,  then  if  P  be  any 
point  on  the  semicircle  or  major  arc  a  +  6,  take 
*D'C'P'=*DCP  with    P'   on   the   same   side  of 


A  SECT  CALCULUS. 


93 


D'O  with  reference  to  A'  as  P  of  DC  with  reference 
to  A.     Thus   D'P'=DPt   also    ^DPA^^D'P'A' 


Fig.  91. 

and  t(DAP  =  $D'A'P',  since  ,4  and  A'  are  on  the 
same  side  of  DP  and  D'P'.     :.  aAPD=  aA'P'D'. 

218.  Definition.  If  an  angle  at  the  center  is 
right  the  arc  it  intercepts  is  called  a  quadrant. 

219.  Corollary  to  217.  In  a  circle  or  in  circles 
with  equal  radii  the  sum  of  any  two  quadrants  is 
congruent  to  the  sum  of  any  other  two,  and  all 
semicircles  are  congruent. 

A  circle  is  the  sum  of  two  semicircles  or  four 
quadrants. 

Congruent  major  arcs  are  the  sums  of  con- 
gruent semicircles  and  congruent  minor  arcs. 

220.  Convention.  We  may  look  upon  a  semi- 
circle as  an  arc  whose  chord  is  a  diameter,  and 
we  may  look  upon  a  whole  circle  as  a  major  arc 
whose  two  end-points  coincide.  The  explemental 
minor  arc  will  then  be  one  single  point. 

We  may  even  think  of  arcs  on  a  circle  greater 
than  the  whole  circle.  In  such  a  case  certain 
points  on  the  circle  are  considered  more  than  once. 

221.  Any  arc  may  now  be  expressed  as  a  sum 
of  a  number  of  quadrants  and  a  minor  arc. 


94  RATIONAL    GEOMETRY. 


The  Sum  of  Angles. 


222.  Definition.  The  sum  of  two  acute  angles 
or  of  a  right  angle  and  an  acute  angle  is  the  angle 
obtained  by  setting  off  one  against  the  other  from 
its  vertex  with  no  interior  point  in  common,  and 
then  omitting  the  common  ray. 

The  sum  of  any  two  or  more  angles  is  an  aggre- 
gate of  right  angles  and  one  acute  angle  such  that, 
taken  as  angles  at  the  center  of  any  one  circle,  the 
sum  of  the  intercepted  quadrants  and  the  arc  inter- 
cepted by  the  acute  angle  equals  the  sum  of  the 
arcs  intercepted  by  the  angles  to  be  added  together. 

223.  Corollary  to  79.  The  sum  of  the  three 
angles  of  any  rectilineal  triangle  is  two  right  angles. 

The  sum  of  two  supplemental  angles  is  two  right 
angles. 

224.  In  the  familiar  terminology  of  motion 
circles  with  equal  radii  are  called  congruent,  and 
we  say  they  can  be  made  to  coincide  if  the  center 
of  one  be  placed  on  the  center  of  the  other. 

Since,  in  their  congruence,  any  one  given  point  of 
the  one  can  be  mated  with  any  point  of  the  other,  we 
say,  after  coincidence  the  second  circle  may  be  turned 
about  its  center,  and  still  coincide  with  the  first. 

Hence  also  a  circle  can  be  made  to  slide  along  itself 
by  being  turned  about  its  center.  This  expresses  a 
fundamental  characteristic  of  the  circle.  It  allows 
us  to  turn  any  figure  connected  with  the  circle  about 
the  center  without  changing  its  relation  to  the  circle. 
Such  displacement  is  called  a  rotation. 

A   displacement   of   a   figure   connected   with    a 


A  SECT   CALCULUS. 


95 


straight,  in  which  the  straight  slides  on  its  trace, 
is  called  translation. 

That  translation  can  be  effected  without  rotation 
is  an  assumption  about  equivalent  to  the  parallel 
Assumption  IV. 

225.  Theorem.  The  diameter  perpendicular  to  a 
chord  bisects  the  angle  at  the  center,  and  the  two 
arcs,  minor  and  major,  made  by  the  chord. 

226.  Convention.  Parallel  secants  or  parallel 
chords  are  said  to  intercept  the  two  arcs  whose 
points  are  between  the  parallels. 

227.  Theorem.  Parallel  chords  intercept  congruent 
arcs. 

Given  AB\\A'B'. 

To  prove  minor  AAr  =  minor 
arc  BB'. 

Proof.  If  CD  J_  AB  then  also 
(by  74)  CD±A'B'.  Then  (by 
117  and  58)  ^ACD^^BCD 
and  t(.A'CD  =  ^B'CD.  ;.  (by 
49)  ^ACA'^^BCB'.  :.  (by 
185)  minor  arc  AAf  =  minor  arc  BB' . 

228.  Theorem.  If  a  simple  plane  polygon  be  cut 
into  triangles  by  diagonals  within  the  polygon  the  sum 
of  their  angles,  together  with  four  right  angles,  equals 

twice  as  many  right  angles  as  the 
polygon  has  sides. 

Proof.      By  a    diagonal  within 
the   polygon   cut   off    a    triangle. 
This    diminishes   the    number    of 
sides  by  one  and  the  sum  of  the 
angles  by  two  right   angles.     So  reduce  the  sides 


Fig.  93. 


96  RATIONAL   GEOMETRY. 

to  three.     We  have  left  two  more  sides  than  pairs 
of  right  angles. 

229.  Definition.  The  exterior  angle  at  any  ver- 
tex of  a  polygon  is  the  angle  between  a  side  and  the 
ray  made  by  producing  the  other  side  through  the 
vertex. 

230.  Theorem.  In  any  convex  plane  polygon  the 
sum  of  the  exterior  angles,  one  at  each  vertex,  is  four 
right  angles. 

Proof.  The  exterior  angle  is  the  supplement  of 
the  adjacent  angle  in  the  polygon.  This  pair  gives 
a  pair  of  right  angles  for  every  side.  But  (by  228) 
the  angles  of  the  polygon  give  a  pair  of  right  angles 
for  every  side  except  two. 

Ex.  205.   In  r'tA,  he  makes  £s  =  to  a  and  /?. 

Ex.  206.  Always  mc<%(a-\-b). 

Ex.  207.  From  point  without  acute  #!  a,  is  to  sides 
make   ^  =  a. 

Ex.  208.  The  sect  joining  the  bisection-points  of  the 
non-||  sides  of  a  trapezoid  is  |  to  the  ||  sides  and  half  their 
sum. 

Ex.  209.  How  many  sides  has  a  polygon,  the  sum  of 
whose  interior  ^s  is  double  the  sum  of  its  exterior  ^s? 

Ex.  210.  How  many  sides  has  a  regular  polygon,  four 
of  whose  ^s  are  together  7  r't  ^s? 

Ex.  211.  The  trisection-points  of  the  sides  of  an  equi- 
lateral  A  form  a  regular  hexagon. 

Ex.  212.  The±s  from  A  and  B  upon  mc  are   =. 

Ex.  213.  Find  the  sum  of  3  non-consecutive  £s  of  an 
inscribed  hexagon. 

Ex.  214.  Construct  •!•  A  from  b  and  a+hb',  from  fi 
(or  hb)  and  perimeter. 

Ex.  215.  The  sum  of  the  three  sects  from  any  point 
within  a  A  to  the  vertices  is  <  the  sum  and  >  J  sum 
of  the  3  sides. 


A  SECT  CALCULUS.  97 

Ex.  216.  Construct  -l-r't  A  from  b  +  c. 

Ex.  217.  In  a  given  st'  find  a  point  to  which  sects  from 
2  given  points  have  the  least  sum. 

Ex.  218.  The  sum  of  the  medians  in  A  is  <  the  sum 
and  >  £  sum  of  sides. 

Ex.219.  Construct  A  from  a  — /?  and  c.  a  from  a,  /?, 
r;     A  from  a,  /?,  R. 

Ex.  220.  The  sum  of  2  opposite  sides  of  a  circumscribed 
quad'  is  half  the  perimeter.  The  sum  of  the  £s  they  sub- 
tend at  the  center  is  2  r't  ^s. 

Ex.  22i.  In  r't  A,  a  +  b  =c+2r  =2R  +  2r. 

Ex.  222.  From  the  vertices  of  A  as  centers  find  3  radii 
which  give  Os  tangent,  two  and  two. 

Ex.  223.  If  H  is  orthocenter,  the  4  circum-Os  of  A,  B, 
C,  H  are   =. 

Ex.  224.  Of  /,  /„  72»  ^3»  each  is  the  orthocenter  of  the 
other  3 ,  and  the  4  circum-  0  s  are  m . 

Ex.  225.  If,  of  a  pentagon,  the  sides  produced  meet, 
the  sum  of  the  ^s  formed  is  2  r't  ^s. 

Fx.  226.  If  hb  meets  b  at  D,  construct  A  from  ho,  a— AD, 
c-CD. 

Ex.  227.  A  quad'  is  a  trapezoid  if  an  opposite  pair  of 
the  4  As  made  by  the  diagonals  are    ■. 


CHAPTER  IX. 


PROPORTION  AND  THE  THEOREMS  OF  SIMILITUDE. 


231.  With  help  of  the  just-given  sect-calculus 
Euclid's  theory  of  proportion  can  in  the  following 
manner  be  established  free  from  objection  and  with- 
out the  Archimedes  assumption. 

232.  Convention.  If  a,  b,  a',  bf  are  any  four  sects, 
then  the  proportion  a:b=a' :  b'  shall  mean  nothing 
but  the  sect  equation  ab'  =ba'. 

233.  Definition.  Two  triangles  are  called  similar 
if  their  angles  are  respectively  congruent.  Sides 
between  vertices  of  congruent  angles  are  called  cor- 
responding. 

234.  Theorem.  In  similar  triangles  the  sides  are 
proportional. 

corresponding  sides  in  two 
similar  triangles. 

To  prove  the  proportion 
a:b=a':b'. 

Proof.    We  consider  first 
the    special     case,    where 
the  angles  included  by  a, 
b  and  a',  b'  in  the  two  tri- 
Fig.  94.  angles  are  right,  and  sup- 

pose both  triangles  on  the  same  right  angle.      We 


Given 


PROPORTION  AND    THE    THEOREMS   OF  SIMILITUDE.    99 


then  set  off  from  the  vertex  on  one  side  the  sect  i, 
and  take  through  the  end-point  of  sect  i  the  parallel 
to  the  two  hypothenuses.  This  parallel  determines 
on  the  other  side  the  sect  e.  Then  is,  by  our  defini- 
tion of  the  sect-product,  b=ea,  b'=ea'.  Conse- 
quently we  have  ab'=ba\  that  is,  a:b=a' :  b'. 

We  pass  now  to  the  general  case.  Construct  in 
each  of  the  two  similar  triangles  the  in-center  /, 
respectively  /',  and  drop 
from  these  the  three  per- 
pendiculars r,  respect- 
ively r',  on  the  triangle's 
sides.  Designate  the  re- 
spective sects  so  deter- 
mined on  the  sides  of  the 
triangles  by  ab}  ac,  bCi  ba, 


Fig.  95. 


cb,  respectively  ab,  a/,  b/t  bc 


r  '     rJ 


The  just- 


proven  special  case  of  our  theorem  gives  then  the 
proportions 

ab:r  =  ab :  r',      bc:r=b/:r', 


ac:r=ac':  r', 


ba:r 


Wir*. 


From  these  we  conclude  by  the  distributive  law 

a\r  =  a':r'y        b\r=b':r\ 

and  consequently,   in   virtue  of  the  commutative 
law  of  multiplication, 

a:  a'  =r:r'  =b:b',     and     a:b  =  a':b'. 

235.  From  the  just-proven  theorem  (234)  we  get 
easily  the  fundamental  theorem  of  the  theory  of 
proportion,  which  is  as  follows: 


ioo  RATIONAL   GEOMETRY. 

Theorem.  If  two  parallels  cut  off  on  the  sides  of 
any  angle  the  sects  a,  b,  respectively  a',  b\  then 
holds  good  the  proportion  a:  b=a' ':  b' .  Inversely, 
when  four  sects  a,  b,  a',  b'  fulfill  this  proportion,  if 
the  pairs  a,  a'  and  b,  b'  are  set  off  upon  the  respective 
sides  of  any  angle,  then  the  straight  joining  the  end- 
points  of  a  and  b  is  parallel  to  that  joining  the  end- 
points  of  a'  and  bf . 

Proof.  First,  since  parallels  make  with  the  sides 
of  the  given  angle  similar  triangles, 
therefore  (by  234)  a:  b=a' :  bf . 

Second,  for  the  inverse.  Through 
the  end-point  of  a!  draw  a  parallel 
to  the  straight  joining  the  end- 
points  of  a  and  b,  and  call  the  sect 
it  determines  on  the  other  side  b" . 
Then  by  First  a:b=a':  b" .  But 
by  hypothesis  a\b=a' :bf .     .'.  b"  =b'. 

236.  Thus  we  have  founded  with  complete  rigor 
the  theory  of  proportion  on  the  basis  of  the 
Assumption-groups  I-IV. 

237.  Corollary  to  235.  If  straights  are  cut  by 
any  number  of  parallels  the  corresponding  inter- 
cepts are  proportional. 

238.  Corollary  to  234.  Parallels  are  divided 
proportionally  by  any  three  copunctal  transversals. 

239.  Corollary  to  235.  Two  triangles  are  similar 
if  they  have  two  sides  proportional  and  the  in- 
cluded angles  congruent. 

240.  Definition.  A  point  P,  costraight  with  A B, 
but  without  the  sect  AB,  is  said  to  divide  the 
sect  AB  externally  into  the  sects  PA,  PB. 


7 

\b' 

V 
Fig.  96. 

PROPORTION  AND   THE    Tk£GR£M$  'OF  ZIMJUT-'JDE.    101 

241.  Corollary  to  235.  A  sect  can  be  divided 
nternally  or  externally  in  proportion  to  any  two 
unequal  given  sects.  The  point  of  internal  divi- 
sion is  unique;  likewise  the  point  of  external  divi- 
sion. 

242.  Theorem.     The  bisector  of  any  angle  of  a 


Fig.  97. 


triangle  or  of  its  adjacent  angle  divides  the  opposite 
side  in  proportion  to  the  other  two  sides. 

[Proof.  Take  AF  ||  to  bisector  BD.  Then 
BF  =  c] 

243.  Definition.  A  sect  divided  internally  and 
externally  in  proportion  is  said  to  be  divided  har- 
monically, and  the  four  points  are  called  a  harmonic 
range. 

244.  Theorem.  A  perpendicular  from  the  right 
angle  to  the  hypothenuse  divides  a  right-angled  tri- 
angle into  two  others  similar  to  it,  and  is  the  mean 
proportional   between   the  parts   of   the  hypothenuse. 

Each  side  is  the  mean  propor- 
tional between  the  hypothenuse  and 
its  adjoming  part. 

Proof.      The    r't    a  ABC    ~    r't  c 
&ABD,  since   ^A  is  common. 


Fig.  98. 


IQ2  gJTIQNAl   GEOMETRY. 

©245.  Corollary    to    244.       The 
perpendicular    from    any    point 
a  in  a  circle  to  the  diameter  is  the 
mean   proportional  between  the 
parts  of  the  diameter. 

Fig.  99. 

246.  Theorem.     The    square    of    the    hypothenuse 
equals  the  sum  of  the  squares  of  the  two  sides. 

Proof.    AC :  AB  =AB  :AD,  that  is,  AB2  =AC-AD. 
Same  way  BC2=AC-DC.     Now  add. 

.'.  AB*  +  BC2=AC(AD  +  DC)=AC2. 

247.  Theorem.     Triangles  having  their  sides  taken 
in  order  respectively  proportional  are  similar. 


Fig.  100. 

In  the  triangles  ABC  and  A'B'C  let  AB:A'B'  = 
AC:A'C=BC:B'C'. 

To  prove  that  the  triangles  ABC  and  A'B'C 
are  similar  (~). 

Proof.  Upon  AB  take  AF=A'B\  and  upon 
AC  take  AH=A'C.  Then  AB:AF=AC:AH. 
.'.(by  239)  a  ABC  ~  to  aAFH.  .\AB:AF  = 
BC:FH.  But  by  hypothesis  AB'.AF  =  BC:B'C. 
:.FH=B'C.  :.  aAFH=aA'B'C  [3  sides  s]. 
.-.  a  ABC-  a  A'B'C. 


PROPORTION  AND   THE   THEOREMS   OF  SIMILITUDE.    103 

248.  Theorem.  The  product  of  the  sects  into 
which  a  given  point  divides  chords  of  a  given  circle 
is  constant. 


Fig.   ioi. 

Hypothesis.  Let  chords  AB  and  CD  intersect 
in  P. 

Conclusion.     AP-PB=CP-PD. 

Proof.     lPAC=tPDB  (by  133),  and 

*    tAPC=^BPD\    \\  aAPC-aBPD. 

249.  Corollary  to  248.  From  a  point  taken  on  a 
tangent  the  square  on  the  sect  to  the  point  of  con- 
tact equals  the  product  of  the  sects  made  on  any 
secant. 


The  Golden  Section. 

250.  Problem.  To  divide  a  sect  so  that  the  product 
of  the  whole  and  one  part  equals  the  square  of  the 
other  part. 

Required  on  AB  to  find  P  such  that  AB-PB  = 
AP\ 

Construction.  Draw  BC  LB  A  and  =\AB  On 
the  straight  AC  take  D  between  A  and  C,  and  E 
beyond  C  such  that  CD^CB=  CE.     Take  AP=AD 


io4 


RATIONAL    GEOMETRY. 


and   AP'=AE.     P   and   P'    divide   AB    internally 
and  externally  in  the  golden  section. 

Proof.     By    249,    AB2=AD-AE=AP(AP  +  AB) 

=  AP2  +  AP-AB. 
.-.  AB(AB  -AP)  =AP2.      .\  AB-PB  =AP\ 
Again,  AB2  =  AE-AD=P'A(AE-DE) 

=  P'A{P,A-AB)=P'A2-AB-P'A. 
.\AB(AB  +  P'A)=P'A2. 
;.AB-P'B  =  P'A2. 


p' 


Fig.   102 
251.  Corollary  to  250.     If  a  is  any  sect  divided 


in 


the  golden  section,  its  greater  part  #=-f(5)*  —  I J 


For  (by  246)   AC2  =  AB2+BC2=a2+-  = 
1  v  J  '  4     \    2 

.-.  AP=AD  =  AC-CD=ia(5)*--- 

252.  Theorem.     The    products    of    opposite    sides 
A  of  a   non-cyclic  quadrilateral 

are  together  greater  than   the 
product  of  its  diagonals. 

Proof.      Make      ifBAF^ 
•4  CAD,       and        ^ABF^ 
4-ACD.     Join  FD. 
Then  aABF-  aACD, 
:.BA:AC=FA-AD. 


Fig.  103. 


PROPORTION  AND    THE    THEOREMS   OF  SIMILITUDE.     105 

But  this  shows  (since  4BAC=^FAD), 
aBAC^aFAD. 

From     aABF-aACD,     .'.  AB-CD  =  BF -AC. 

From       aBAC-aFAD,     .-.  BC-AD  =FD-AC. 
?.  AB-CD  +  BC-AD=BF'AC  +  FD-AC>BD-AC. 

253.  Corollary  to  252  (Ptolemy).  The  product 
of  the  diagonals  of  a  cyclic  quadrilateral  equals 
the  sum  of  the  products  of  the  opposite  sides. 
(For  then  F  falls  on  BD.) 

254.  Definition.  Similar  polygons  are  those  of 
which  the  angles  taken  in  order  are  respectively 
equal  [i.e.,  congruent],  and  the  sides  between  the 
equal  angles  proportional. 

255.  Theorem.  Two  similar  polygons  can  be 
divided  into  the  same  number  of  triangles  respect- 
ively similar. 

256.  Theorem.  If  a  cyclic  polygon  be  equilateral 
it  is  regular. 

Ex.  228.  If  AB  is  divided  harmonically  by  P,  P',  then 
PP'  is  divided  harmonically  by  A,  B. 

Ex.  229.  If  two  triangles  have  the  sides  of  one  respect- 
ively parallel  or  perpendicular  to  the  sides  of  the  other 
they  are  similar. 

Ex.  230.  The  corresponding  altitudes  of  two  similar  tri- 
angles are  proportional  to  any  two  corresponding  sides. 

Ex.  231.  To  divide  a  sect  into  parts  proportional  to  given 
sects. 

Ex.  232.  A  sect  can  be  divided  into  any  number  of 
equal  parts. 

Ex.  233.  To  find  the  fourth  proportional  to  three  given 
sects. 

Ex.  234.  To  find  the  third  proportional  to  two  given  sects. 

Ex.  235.  If  three  non-parallel  straights  intercept  pro- 
portional sects  on  two  parallels  they  are  copunctal. 


106  RATIONAL    GEOMETRY. 

Ex.  236.  Every  equiangular  polygon  circumscribed 
about  a  circle  is  regular. 

Ex.  237.  Every  equilateral  polygon  circumscribed  about 
a  circle  is  regular  if  it  has  an  odd  number  of  sides. 

Ex.  238.  Every  equiangular  polygon  inscribed  in  a 
circle  is  regular  if  it  has  an  odd  number  of  sides. 

Ex.  239.  One  side  of  a  A  is  to  either  part  cut  off  by  a 
st'  ||  to  the  base  as  the  other  side  is  to  the  corresponding 
part. 

Ex.  240.  If  a  straight  divides  two  sides  of  a  A  propor- 
tionally, it  is  ||  to  the  third  side. 

Ex.  241.  The  bisectors  of  an  interior  and  an  exterior  4- 
at  one  vertex  of  a  A  divide  the  opposite  side  harmonically. 

Ex.  242.  The  perimeters  of  two  ~  polygons  are  pro- 
portional to  any  two  corresponding  sides. 

Ex.  243.  A  median  and  two  sides  of  a  trapezoid  are 
copunctal. 

Ex.  244.  The  chords  on  a  st'  through  a  contact-point 
of  two  Os  are  proportional  to  their  diameters;  and  a 
common  tangent  is  a  mean  proportional  between  their 
diameters. 

Ex.  245.  The  sum  of  the  squares  of  the  segments  of 
2  J_  chords  equals  the  sq'  of  the  diameter. 

Ex.  246.  On  the  piece  of  a  tangent  between  two  ||  tan- 
gents the  contact-point  makes  segments  whose  product 
is  the  square  of  the  radius. 

Ex.  247.  To  inscribe  in  and  circumscribe  about  a  given 
O  a  A<v  toa  given  A. 

Ex.  248.  The  hypothenuse  is  divided  harmonically  by 
any  pair  of  st's  through  the  vertex  of  the  r't  if  making 
m  4^s  with  one  of  its  sides. 

Ex.  249.  The  bisection-point  of  the  base  of  a  A  and  any 
point  on  a  ||  to  the  base  through  the  vertex  make  a  sect 
cut  harmonically  by  a  side  and  the  other  side  produced. 

Ex.  250.  /  divides  h  as  b  to  a~\-c. 

Ex.251.  Rriac  =b:2(a-\-b  +  c). 

Ex.  252.  In  a  r't  A  the  X  sides  are  as  the  in-radii  of 
As  made  by  he. 


PROPORTION  AND  THE  THEOREMS  OF  SIMILITUDE.  107 

Ex.  253.  Sects  from  the  ends  of  the  base  of  a  a  to  the 
intersections  of  a  ||  to  base  with  the  sides  intersect  on  a 
median. 

Ex.  254.    A   from  ft,  a/c,  R. 

Ex.  255.  A  quad'  is  cyclic  if  diagonals  cut  so  that 
product  of  segments  of  one  equals  product  of  segments 
of  the  other. 

Ex.  256.  The  sides  of  the  pedal  cut  off  as  ~  to  the 
original. 

Ex.  257.  Three  points  being  given,  to  determine  an- 
other, through  which  if  any  st'  be  drawn,  Is  upon  it  from 
two  of  the  former,  shall  together  be  equal  to  the  ±  from 
the  third. 

Ex.  258.  From  two  given  sects  to  cut  off  two  propor- 
tional to  a  second  given  pair  so  as  to  leave  remainders 
proportional  to  another  given  pair. 

Ex.  259.  If  one  chord  bisect  another,  and  tangents  from 
the  extremities  of  each  meet,  the  st'  of  their  intersection 
points  is  ||   to  the  bisected  chord. 

Ex.  260.  In  A,  if  sects  from  the  ends  of  the  base  to  the 
opposite  sides  intersect  on  the  altitude,  the  joins  of  its 
foot  to  their  ends  will  make  equal  angles  with  the  base. 

Ex.  261.  The  diagonals  of  a  regular  pentagon  cut  one 
another  in  the  golden  section,  and  the  larger  segments 
equal  the  sides. 

Ex.  262.  From  the  vertex  of  an  inscribed  A  a  sect  to 
the  base  ||  to  a  tangent  at  either  end  of  the  base  is  a  fourth 
proportional  to  the  base  and  two  sides. 

Ex.  263.  Straights  from  the  vertices  of  any  a  to  the 
contact-points  of  the  in-o  are  copunctal. 

Ex.  264.  Construct  a  from  b,  ft,  and  that  tb  makes 
segments  as  m  to  n. 

Ex.  265.   A  from  ft,  mb,  and  •£  between  6  and  nib. 

Ex.  266.    A  from  tb  and  J_s  on  it  from  A  and  C. 

Ex.  267.  A  from  ft,  a  —  c,  and  difference  of  segments 
made  by  hb. 

Ex.  268.   R't  A  from  a  +  b,  and  b  +  c. 

Ex.  269.    a   from   a -ft,  a:b—ni:n,  and  a  third  propor- 


Io8  RATIONAL    GEOMETRY. 

tional  to  the  difference  of  segments  made  by  hc  and  the 
lesser  side. 

Ex.  270.    a  from  a,  a  +  b,  a-\-c. 

Ex.  271.    a  from  a,  R,  and  b:c  =m:n. 

Ex.  272.    a  from  a,  b,  a—hb. 

Ex.  273.  Divide  a  given  sect  harmonically  as  m  to  n. 

Ex.274.  In~AS,  a:af  =ha:hfa  =ma:m'a  =ta:t'a  =r:r'  = 
R:R'. 

Ex.  275.  Two  r't  As  are  **»  if  hypothenuse  and  a  J_  are 
proportional. 

Ex.  276.  If  a  chord  is  bisected  by  another,  either  seg- 
ment of  the  first  is  a  mean  proportional  between  the  seg- 
ments of  the  other. 

Ex.  277.  R't  A  from  a  and  the  non-adjacent  segment 
made  by  hc. 

Ex.  278.  The  diameter  of  a  0  is  a  mean  proportional 
between  the  sides  of  the  circumscribed  regular  A  and. 
hexagon. 

Ex.  279.  From  the  center  of  a  given  O  to  draw  a  st' 
cutting  off  from  a  given  tangent  a  sect  any  multiple  of 
the  segment  between  O  and  tangent. 

Ex.  280.  If  2  As  have  two  sides  of  the  one  proportional 
to  two  sides  of  the  other,  and  ^  s,  one  in  each,  opposite 
one  corresponding  pair  of  these  sides  = ,  the  ^  s  opposite 
the  other  pair  are  either  =  or  supplemental. 

Ex.  281.  The  altitude  to  hypothenuse  is  a  fourth  pro- 
portional to  it  and  the  sides. 

Ex.  282.  The  vertices  of  all  as  on  the  same  base  with 
sides  proportional  are  on  a  O  with  center  costraight  with 
base  and  radius  a  mean  proportional  between  the  sects 
from  its  center  to  the  ends  of  base. 

Ex.  283.  To  inscribe  a  sq.  in  a  given  A. 


CHAPTER  X. 

EQUIVALENCE. 

The  theory  of  equivalence  in  the  plane. 

257.  We  take  as  basis  for  the  investigations  in  the 
present  chapter  the  Assumptions  I,  1-2,  and  II-IV. 
We  exclude  the  Archimedes  assumption.  Our 
theory  of  proportion  and  sect-calculus  put  us  in 
position  to  found  the  Euclidean  theory  of  equiva- 
lence by  means  of  the  assumptions  named,  that  is, 
in  the  plane  and  independent  of  the  Archimedes 
assumption. 

258.  Convention.  If  we  join  two  points  of  a 
polygon  P  by  any  sect-train  which  runs  wholly  in 
the  interior  of  the  polygon  we  obtain  two  new  poly- 
gons, Pl  and  P2,  whose  inner  points  all  lie  in  the 
interior  of  P. 

We  say:  P  is  separated  or  cut  into  Px  and  P2; 
Pt  and  P2  together  compose  P. 

259.  Definition.  Two  polygons  are  called  equiv- 
alent if  they  can  be  cut  into  a  finite  number  of 
triangles  congruent  in  pairs. 

260.  Definition.  Two  polygons  are  said  to  be 
equivalent  by  completion  if  it  is  possible  so  to  annex 

IOQ 


no 


RATIONAL    GEOMETRY. 


equivalent  polygons  to  them  that  the  two  polygons 
so  composed  are  equivalent. 

261.  We  will  use  the  sign  of  equality  (  =)  between 
polygons  to  denote  * '  equivalent  by  completion. ' ' 

262.  From  these  definitions  follows  immediately: 
By  uniting  equivalent  polygons  we  get  again  equiv- 
alent polygons.  If  we  take  away  equivalent  poly- 
gons from  equivalent  polygons  the  remaining  poly- 
gons are  equivalent  by  completion. 

Furthermore,  we  have  the  following  propositions : 

263.  Theorem.  Two  polygons  P,  and  P2  equiv- 
alent to  a  third  P3  are  equivalent. 

Two  polygons  equivalent  by  completion  to  a  third 
are  equivalent  by  completion. 

Proof.  By  hypothesis  there  is  as  well  for  Px  as 
for  P2  an  assignable  partition  into  triangles  such 
that  each  of  these  two  partitions  corresponds  re- 
spectively to  a  partition  of  the  polygon  P3  into 
congruent  triangles. 


\             • 

\ 

1          \ 

\^ 

\          / 
\      / 

Fig.  104. 

If  we  consider  these  two  partitions  of  P3  simul- 
taneously every  triangle  of  the  one  partition  will  in 


EQUIVALENCE.  HI 

general  be  cut  into  polygons  by  sects  which  pertain 
to  the  other  partition.  Now  we  introduce  a  suffi- 
cient number  of  sects  to  cut  each  of  these  polygons 
itself  into  triangles,  and  then  make  the  two  corre- 
sponding partitions  into  triangles  in  P,  and  in  P2. 
Then  these  two  polygons  Px  and  P2  are  cut  into  the 
same  number  of  triangles  congruent  in  pairs,  and 
are  therefore  by  our  definition  equivalent. 

Again,  if  Q,  =  Q3  and  Q2  =  Q3,  then  according  to 
definition  the  composite  Ql  +  P1  is  equivalent  to 
Q3  +  Plt  and  Q2  +  P2  is  equivalent  to  Q3  +  P2.  There- 
fore Qx  +  Px  +  P2  is  equivalent  to  Q3  +  Px  +  P2,  which 
is  equivalent  to  Q2  +  P2  +  Pi-     •'•  Qi=02- 

Parallelograms  and  Triangles  with  equal  bases  and 
altitudes. 

264.  Theorem.  Two  parallelograms  with  equal 
bases  and  equal  altitudes  are  equivalent  by  com- 
pletion. 

Proof.  aBAE=aCDF.  Annex  aBCH  and 
leave  out  a  DUE.     ; .  ABCD  =  EBCF. 


Fig.   1P5.  Fig.   106. 

To  prove  these  parallelograms  equivalent  would 
require  here  the  Archimedes  assumption. 


H2  RATIONAL   GEOMETRY. 

265.  Theorem.  Any  triangle  ABC  is  always 
equivalent  to  a  certain  parallelo- 
gram with  equal  base  and  half 
altitude. 

Proof.  Bisect  AC  in  D  and  BC 
in  E,  and  then  prolong  BE  to  F, 
making  EF  =  DE.  The  triangles 
DEC  and  FBE  are  then  congruent, 
and  consequently  the  A  ABC  and 
parallelogram  ABFD  are  equivalent. 

266.  From  264  and  265  follows  with  help  of 
263  immediately: 

Theorem.  Two  triangles  of  equal  bases  and  equal 
altitudes  are  equivalent  by  completion. 

267.  That  two  triangles  with  equal  bases  and 
altitudes  are  always  equivalent  cannot  possibly 
be  proven  without  using  the  Archimedes  assump- 
tion. 

268.  The  remaining  theorems  of  elementary 
geometry  about  the  equivalence  by  completion  of 
polygons,  and  also,  in  particular,  the  Pythagoras 
equivalence  theorem :  ' '  The  square  on  the  hy- 
pothenuse  of  a  right  triangle  is  equivalent  to  the 
united  squares  on  the  other  two  sides,"  are  easy 
consequences  of  the  theorems  just  set  up. 

269.  But,  nevertheless,  in  further  working  out 
the  theory  of  equivalence  we  encounter  an  essen- 
tial difficulty. 

In  particular  our  considerations  hitherto  leave 
undecided  whether  perhaps  all  polygons  are  not 
always  equivalent  by  completion  to  one  another. 
In  this    case   all  the    previously  established    theo- 


EQUIVALENCE.  113 

rems  would  teach  nothing  and  be  without  import- 
ance. 

The  proven  theorems  about  equivalence  by  com- 
pletion are  entirely  rigorous;  nevertheless  we 
recognize  on  closer  investigation  that  they  all  for 
the  present  have  no  content.  We  do  not  yet  know 
whether  there  are  polygons  at  all  which  are  not 
equivalent  by  completion. 

270.  And  not  only  must  we  know  this,  if  we 
would  undertake  anything  with  our  theorems,  but 
also  we  need  to  consider  the  more  specific  question 
whether  two  rectangles  equivalent  by  completion, 
having  one  common  side,  have  also  necessarily 
their  other  sides  congruent,  that  is,  whether  a 
rectangle  is  uniquely  determined  by  one  of  its 
sides  and  its  equivalence  by  completion. 

271.  As  the  closer  consideration  shows,  we  need 
for  answering  the  questions  raised  the  inverse  of 
266,  which  runs  as  follows: 

Theorem.  If  two  triangles  equivalent  by  completion 
have  equal  bases  then  they  have  also  equal  altitudes. 

This  fundamental  theorem  is  the  thirty-ninth 
of  the  first  book  of  Euclid's  Elements  (Eu.  I,  39). 
However,  to  prove  it  Euclid  invokes  the  general 
theorem  about  magnitudes :  ' '  The  whole  is  greater 
than  its  part,"  a  procedure  which  amounts  to  the 
introduction  of  a  new  geometric  assumption,  that 
is,  the  tacit  assumption  of  a  new  and  independent 
magnitude,  the  "  surface"  or  u  superficial  content." 

272.  Now  for  the  question  of  superficial  content, 
we  can,  on  the  basis  of  only  our  old  assumptions, 
though  into  them  the  word  M content"  does  not  in 


H4  RATIONAL   GEOMETRY. 

any  way  enter,  prove  that  two  polygons  can  be 
compared  as  to  content. 

273.  Thus  the  congruence  and  equality  of  sects 
is  fundamental  or  primitive,  rooted  immediately 
in  assumptions. 

274.  But  the  equality  of  polygons  as  to  content 
is  a  constructible  idea  with  nothing  new  about  it 
but  a  definition. 

275.  We  proceed  now  to  establish  this  theorem 
(Eu.  I,  39)  and  therewith  the  theory  of  content 
in  the  way  we  desire,  that  is,  merely  with  help  of 
the  plane  assumptions  without  using  the  Archimedes 
assumption. 

276.  It  need  not  surprise  us  that  the  proof  is  not 
wholly  simple.  For  that  two  triangles  are  equiva- 
lent by  completion  according  to  definition  only  says 
that  certain  "corresponding"  triangle-partitions 
exist;  thereby  can  the  number  of  the  triangles  be 
very  great  and  one  does  not  immediately  see  how 
from  that  we  can  conclude  from  equality  of  bases 
equality  of  the  altitudes. 

277.  We  begin  by  introducing  the  idea  of  area. 

The  area  of  triangles  and  polygons. 

278.  Definition.  In  any  triangle  ABC  with  the 
sides  a,  b,  c,  if  we  construct 
the  two  altitudes  ha=AD, 
hb=BEy  then  follows  from 
the  similarity  of  the  tri- 
angles BCE  and  A  CD, 
(by  234)  the  proportion 
a:hb  =  b:ha,  that  is,  aha=bhb.  fig.  10s. 


EQUIVALENCE. 


115 


Consequently  in  every  triangle  the  product  of  a 
base  and  its  altitude  is  independent  of  what  side 
of  the  triangle  one  chooses  as  base.  Half  the 
product  of  base  and  altitude  of  a  triangle  a  is 
called  the  area  of  the  triangle  A,  and  designated 
by  A(A). 

279.  Convention.  A  sect  which  joins  a  vertex 
of  a  triangle  with  a  point  of  the  opposite  side  is 
called  a  transversal;  this  cuts  the  triangle  into  two 
triangles  with  common  altitude,  whose  bases  lie 
on  the  same  straight.  Such  a  partition  is  called  a 
transversal  partition  of  the  triangle. 

280.  Theorem.  //  a  triangle  A  is  in  any  way  cut 
by  any  straights  into  a  certain  finite  number  of  tri- 
angles A  kt  then  is  always  the  area  of  the  triangle  a 
equal  to  the  sum  of  the  areas  of  all  the  triangles  a  k. 

Proof.  From  the  distributive  law  in  our  sect- 
calculus  follows  immedi- 
ately that  the  area  of  any 
triangle  is  equal  to  the 
sum  of  the  areas  of  two 
triangles  which  arise  from 
the  first  by  any  transversal 
partition.     Thus,  for  example, 

A(A1)+A(A2)=$b1h  +  ib2h=iHbl  +  b2)  =  ibh  =  A(A). 
The  repeated  application  of  this  fact  shows  that 
the  area  of  any  triangle  is  also 
equal  to  the  sum  of  the  areas  of 
all  the  triangles  which  arise 
from  the  first,  if  we  make  suc- 
cessively however  many  trans- 
versal partitions. 


Fig. 


09. 


n6  RATIONAL   GEOMETRY. 

In  order  now  to  accomplish  the  corresponding 
demonstration  for  any  partition  of  the  triangle  a  into 
triangles  Ajb>  we  draw  from  one  vertex  A  of  the 
triangle  A  through  each  dividing-point  of  the  par- 
tition, that  is,  through  each  vertex  of  the  triangles 
A  *,  a  transversal ;  by  these  transversals  the  triangle 
a  is  cut  into  certain  triangles  A,.  Each  of  these 
triangles  A  t  is  cut  by  the  dividing-sects  of  the  given 
partition  into  certain  triangles  and  quadrilaterals. 
If  in  each  quadrilateral  we  draw  a  diagonal  then 
each  triangle  A  t  is  cut  into  certain  triangles  A  ts. 

We  will  now  show  that  the  partition  into  trian- 
gles Afc,  as  well  for  the  triangles  a,  as  also  for 
the  triangles  A  k,  is  a  chain  of  transversal  partitions. 


Fig.  hi. 

In  fact,  first  is  clear,  that  every  partition  of  a  tri- 
angle into  part-triangles  can  always  be  effected  by 
a  series  of  transversal  partitions,  if,  in  the  partition, 
no  dividing-points  lie  in  the  interior  of  the  triangle, 
and  besides  at  least  one  side  of  the  triangle  remains 
free  from  dividing-points. 

Now  these  conditions  are  evident  for  the  trian- 
gles a  t  from  the  circumstance  that  for  each  of  them 


EQUIVALENCE.  1 1 7 

the  interior  and  one  side,  that  opposite  the  point 
A,  are  free  from  dividing-points. 

But  also  for  every  a^  is  the  partition  into  a/s 
reducible  to  transversal  partitions.  In  fact,  if  we 
consider  a  triangle  a  k,  then  there  is,  among  the 
transversals  from  A  in  the  triangle  a  a  certain 
transversal  which  either  cuts  the  triangle  a  k  into 
two  triangles,  or  else  upon  which  a  side  of  a  k  falls. 
For  we  recall  that  within  no  At  are  there  dividing- 
points. 

By  construction,  through  every  vertex  of  Ajt  goes 
a  transversal  from  A  ;  and  there  is  always  one  vertex 
of  &  k  for  which  this  transversal  has  a  second  point 
not  in  the  region  exterior  to  A  * ;  it  therefore  either 
goes  through  the  interior  of  A  k  or  upon  it  is  a  side 
of  A*. 

In  this  latter  case  this  side  of  the  triangle  Ak 
remains  altogether  free  from  further  dividing-points 
in  the  partition  into  triangles  a/5.  In  the  other 
case  the  sect  of  that  transversal  within  the  triangle 
A  k  is  for  both  the  triangles  thus  arising  a  side  which 
in  the  partition  into  triangles  a  ts  remains  surely  free 
from  further  dividing-points. 

From  the  considerations  at  the  beginning  of  this 
demonstration  the  area  A  (a)  of  the  triangle  a 
equals  the  sum  of  all  areas  A  (a{)  of  the  triangles 
A  t,  and  this  sum  is  equal  to  the  sum  of  all  areas 
A(Ats).  On  the  other  hand  is  also  the  sum  of  the 
areas  A(Ak)  of  all  triangles  a k  equal  to  the  sum  of 
all  areas  A(ais).  Hence,  finally,  the  area  A  (a)  is 
also  equal  to  the  sum  of  all  areas  A(Ak).  So  the 
theorem  is  completely  proven.     « 


n8  RATIONAL   GEOMETRY. 

281.  Definition.  If  we  define  the  area  A(JP)  of 
a  polygon  as  the  sum  of  the  areas  of  all  triangles 
into  which  it  is  cut  in  a  certain  partition,  then  the 
area  of  a  polygon  is  independent  of  the  way  it  is 
cut  into  triangles,  and  consequently  determined 
uniquely  simply  by  the  polygon  itself. 

Proof.  Suppose  a  c  to  be  the  triangles  of  a  cer- 
tain partition,  and  A  k  those  of  any  other  partition. 
Considering  these  two  partitions  simultaneously,  in 
general  is  every  triangle  Ac  cut  into  polygons  by 
sects  pertaining  to  a*.  Now  we  introduce  sects 
sufficient  to  cut  these  polygons  themselves  into  tri- 
angles As.  Then  the  triangles  a  c  have  (by  280)  for 
the  sum  of  their  areas  the  sum  of  the  areas  of  a5. 
But  so  also  have  the  triangles  A  &. 

[This  fact,  that  the  sum  named  is  independent  of 
the  way  of  cutting  up  the  polygon,  is  the  kernel, 
the  essence  of  this  whole  investigation.] 

282.  Corollary  to  281.  Equivalent  polygons  have 
equal  area. 

283.  Moreover,  if  P  and  Q  are  two  polygons  equiv- 
alent by  completion,  then  there  must  be,  from  the 
definition,  two  equivalent  polygons  P'  and  Q',  such 
that  the  polygon  compounded  of  P  and  P1  is  equiv- 
alent to  the  polygon  compounded  of  Q  and  Q' .  From 
the  two  equations 

i4(P  +  P')~.4(Q  +  g'),     A(P')=A(Q')> 
we  deduce  at  once  A(P)  =A(Q),  that  is,  polygons 
equivalent  by  completion  have  equal  area. 

284.  From  this  latter  fact  we  get  immediately 
the  proof  of  the  theorem  of  271  (Eu.  I,  39).     For, 


EQUIVALENCE. 


HO 


designating  the  equal  bases  of  the  two  triangle?, 
by  b,  the  corresponding  altitudes  by  h  and  h\  we 
then  conclude  from  the  assumed  equivalence  by 
completion  of  the  two  triangles  that  they  must 
also  have  equal  area;  that  is,  it  follows  \bh  =  \bh\ 
and,  consequently,  after  division  by  \b,  h=h'\ 
which  was  to  be  proved. 


Area  and  Equivalence-by-completion. 

285.  In  what  precedes  we  have  found  that  poly- 
gons equivalent-by-completion  have  always  equal 
area.     The  inverse  is  also  true. 

286.  To  prove  the  inverse,  we  consider  first 
two  triangles  ABC  and  AB'C  with  a  common 
right  angle  at  A.  The  areas  of  these  two  triangles 
are  expresesd  by  the  formulas 

A(ABC)=iAB.AC, 

A(AB'C'=iAB'-AC. 

If  we  assume  that  these 
two  areas  are  equal,  we 
have 

AB.AC=AB'-AC, 
or  AB:AB'=AC':AC, 
and  from  this  it  follows  (by  235)  that  the  two 
straights  BC  and  B'C  are  parallel,  and  then  we 
recognize  (from  266)  that  the  two  triangles  BOB' 
and  BC'C  are  equivalent-by-completion.  By  an- 
nexing the  triangle  ABC  it  follows  that  the  two 
triangles  ABC  and  AB'C  are  equivalent-by-com- 
pletion.     Thus  we   have   proved   that   two   right- 


FlG.    112. 


120 


RATIONAL    GEOMETRY. 


angled  triangles  with  equal  area  are  also  always 
equivalent-by-completion . 

287.  Take  now  any  triangle,  with  base  b  and 
altitude  h,  then  this  is  (by  266)  equivalent-by- 
completion  to  a  right-angled  triangle  with  the  two 
perpendicular  sides  b  and  h\  and  since  the  original 
triangle  evidently  has  the  same  area  as  the  right- 
angled  triangle,  so  it  follows  that  in  the  preced- 
ing article  the  limitation  to  right-angled  triangles 
was  not  necessary.  Thus  we  have  shown  that 
any  two  triangles  with  equal  area  are  also  always 
equivalent-by -completion. 

288.  Now  let  P  be  any  polygon  with  area  b. 
Let  P  be  cut  into  n  triangles  with  the  respective 
areas  blf  b2,  .  .  .  bn\    then  is  b=b1  +  b2  +  .  .  .  +  bn. 

Construct  now  a  triangle  ABC  with  the  base 
A  B  =b  and  the  altitude  h  =  i  and  mark  on  the  base 
the  points  Alt  A2,  .  .  .  An,  such  that  bl=AAl, 
b2  =A1A2,  .  .  .  bn_x  =i„_2A„.„    bn  =An_1B.      Since 


&n   b 


Fig.   113. 


tne  triangles  within  the  polygon  P  have  respectively 
the  same  area  as  the  triangles  AAlCi  A±A2C,  .  .  .  • 
An_2An_xC,  An_tBC,  so  they  are,  by  what  has  just 
been  proven,  equivalent-by-completion  to  these. 


EQUIVALENCE.  1 2 1 

Consequently  the  polygon  P  is  equivalent-by- 
completion  to  a  triangle  with  the  base  b  and  the 
altitude  h  =  i. 

Hence  follows,  with  help  of  theorem  287,  that 
two  polygons  of  equal  area  are  always  equivalent- 
by-completion. 

289.  We  may  combine  the  two  results  found 
in  this  article  288  and  in  283  into  the  following 
theorem:  Two  polygons  equivalent-by-completion 
have  always  the  same  area;  and  inversely,  two 
polygons  with  equal  area  are  always  equivalent-by- 
completion. 

290.  In  particular  two  rectangles  equivalent-by- 
completion  which  have  one  side  in  common  must 
also  have  their  other  sides  congruent. 

291.  Also  follows  the  theorem:  //  we  cut  a 
rectangle  by  straights  into  several  triangles  and  leave 
out  even  one  of  these  triangles,  then  we  cannot  with  the 
remaining  triangles  fill  out  the  rectangle. 

In  what  precedes  is  shown  that  this  theorem  is 
completely  independent  of  the  iVrchimedes  assump- 
tion. Moreover,  without  the  application  of  the 
Archimedes  assumption,  this  theorem  291  does 
not  suffice  of  itself  for  demonstrating  Eu.  I,  39. 

292.  Definition.  Of  two  polygons  P  and  Q,  we 
call  P  of  lesser  content  (respectively,  of  equal,  of 
greater  content)  than  Qf  according  as  the  area  A(P) 
is  less  (equal,  greater)  than  A(Q). 

293.  From  what  precedes  it  is  clear  that  the 
concepts  of  equal  content,  of  lesser  content,  of 
greater  content  are  mutually  exclusive. 

294.  Further,  we   see  that  a  polygon  which  lies 


122  RATIONAL   GEOMETRY. 

wholly  within  another  polygon  must  always  be  of 
lesser  content  than  this  latter. 

295.  Herewith  we  have  established  the  essential 
theorems  of  the  theory  of  superficial  content, 
wholly  upon  considerations  of  the  congruence  of 
sects  and  angles,  and  without  assuming  superficial 
content  to  be  a  magnitude. 

296.  Theorem.  The  area  of  any  parallelogram  is 
the  product  of  the  base  by  the  altitude. 

297.  Corollary  to  296.  The  area  of  any  rec- 
tangle or  square  is  the  product  of  two  consecutive 
sides. 

298.  A  square  whose  side  is  the  unit  sect  has  for 
area  this  unit  sect, 

since  1X1=1. 

Any  polygon  has  for  area  as  many  such  unit 
sects  as  the  polygon  contains  such  squares  on  the 
unit  sect. 

The  number  expressing  the  area  of  a  polygon 
will  thus  be  the  same  in  terms  of  our  unit  sect  or  in 
terms  of  a  square  on  this  sect  considered  as  a  new 
kind  of  unit,  a  unit  surface,  or  unit  of  content. 
Such  units,  though  traditional,  are  unnecessary 
and  sometimes  exceedingly  awkward,  as,  for  ex- 
ample, the  acre. 

Ex.  284.  If  twice  the  number  expressing  the  area  of  a 
triangle  be  divided  by  the  number  expressing  the  base, 
the  quotient  is  the  number  expressing  the  altitude. 

Ex.  285.  One  side  of  a  triangle  is  35-74,  and  the  alti- 
tude on  it  is  6-3.     Find  the  area. 

299.  Theorem.    //  two  triangles  (or  parallelograms) 


EQUIVALENCE. 


i23 


have  one  angle  of  the  one  congruent  to  one  angle  of  the 
other,  their  areas  are  proportional  to  the  products  of 
their  sides  about  the  congruent  angles. 


then 


Let  4Cm$C% 

Area  a  ABC 


\AC-BB         AC-BD 


Area  &A'B'C~ \A'C -B'U  ~  A'C -B'U' 

R71        BC 
But  in  •  ~  a  's  BCD  and  B'C'D',  -§rjy==g,£r. 

Area  a  ABC        AC-BC 


"  Area  aA'B'C     A'C -B'C 

300.  Corollary  to  299.  The  areas  of  similar  tri- 
angles are  proportional  to  the  squares  of  correspond- 
ing sides. 

301.  Problem.  To  construct  a  rectangle,  given 
two  consecutive  sides. 

Construction.  Take  a  straight  and  a  perpendic- 
ular to  it.     From  the  vertex  of 

? — ---  — -§ 

the  right  angle  take  one  given 
sect  on  the  straight,  the  other   _ 
on  the  perpendicular.     Through 
their    second    end-points    draw 
perpendiculars.     These  (by  77)  meet.     They  inter- 
sect in  the  fourth  vertex  of  the  rectangle  required. 

Proof.  By  construction  the  figure  is  a  parallelo- 
gram with  one  right  angle;   .'.a  rectangle. 


Fig.  115. 


124 


RATIONAL    GEOMETRY. 


302.  Corollary  to  301.     So  we  may  construct  a 
square  on  any  given  sect. 

303.  Theorem.     The  square  on  the  hypothenuse  of 
any  right-angled  triangle  is  equivalent  to  the  sum  of  the 

squares  on  the  other  two  sides. 

Hypothesis,  a  ABC,  r't- 
angled  at  C. 

Conclusion.  Square  on  A B 
is  equivalent  to  sq.  on  AC 
+  sq.  on  BC. 

Proof.  On  hypothenuse 
AB,  on  side  opposite  C,  con- 

ABDF.     From  its  vertices 

,  and    /.  ||  BC.     Drop  BK, 


A 
Il6. 


Fig. 

struct  (by  302)  the  sq. 
D,  Fdrop  DH,  FG±CA 


FL±DH,  and  ?.. 
(complements  of 
(complements  of 
(complements  of 
aDBK=aFDL  = 


||  AC.  Then  $ABC  =  tDBK 
4ABK).  Also  ^BDK^iDFL 
Also  -4-DFL  =  ^AFG 
;.  (by  44),  aABC  = 
.-.  BCHK  is  sq.  on  BC, 
sq.  on  AB  =  AFLKB  + 


^.LDF) 
•4.AFL). 
a  FAG. 
and  FGHL  =sq.  on  AC.  . 
2  A  ABC  =  sq.  on  BC  + sq.  on  AC. 

304.  Problem.     To  construct  an  equilateral  triangle 
on  a  given'  sect. 

Construction.  On  the  st.  AB  from  A  take  the 
given  sect  AB.  At  B  erect  to  AB 
a  perpendicular.  On  this  perpen- 
dicular, from  B  take  BC,  the  given 
sect.  Join  AC.  At  C  erect  to  the 
straight  AC  the  perpendicular  CD. 
On  CD  from  C  take  CD,  the  given 
sect.  Join  AD.  Bisect  AD  at  E, 
"7-  and  AB  at  F.     Erect  at  F  to  AB 


EQUIVALENCE.  125 

the    perpendicular   FG.     From    F   take   FG=AE. 
ABG  is  the  required  equilateral  triangle. 
__Proof.      AG2=AF2  +  FG2  =  (iABy  +  ({3(AB)2  = 
AB\ 

305.  Theorem.  In  any  triangle  y  the  square  of  a 
side  opposite  any  acute  angle  is 

less  than  the  sum  of  the  squares  A^^ 

of  the  other  two    sides  by  twice       y   j*    ^s^ 

the  product  of  either  of  those  sides    CZ — ! i _^A 

and  a  sect  from  the  foot  of  that 
side's  altitude  to  the  vertex  of  the 
acute  angle. 

Proof.  Let  a,  6,  c  denote  the  sides,  and  h  denote 
6's  altitude,  and  /  the  sect  from  its  foot  to  the 
acute  angle  A . 

a2-h2  =  (b-j}2  =  b2-2bj  +  j2  =  b2-2bj  +  c2-h2; 
.'.a2  =  b2-2bj  +  c2. 

306.  Theorem.  In  an  obtuse-angled  triangle  the 
square  of  the  side  opposite  the  obtuse  angle  is 
greater  than  the  sum  of  the  squares  of  the  other 
two  sides  by  twice  the  product  of  either  of  those 
sides  and  a  sect  from  the  foot  of  that  side's  altitude 
to  the  vertex  of  the  obtuse  angle. 

Ex.  286.  Find  the  area  of  an  isosceles  triangle  whose 
base  is  60  and  each  of  the  equal  sides  50. 

Ex.  287.  If  two  triangles  (or  parallelograms)  have  an 
angle  of  one  supplemental  to  an  angle  of  the  other,  their 
areas  are  as  the  products  of  the  sides  including  the  supple- 
mentary angles. 

Ex.  288.  The  area  of  any  circumscribed  polygon  is  half 
the  product  of  its  perimeter  by  the  radius  of  the  inscribed 
circle. 


126  RATIONAL   GEOMETRY. 

Ex.  289.  To  find  the  area  of  a  trapezoid.  Rule: 
Multiply  the  sum  of  the  parallel  sides  (its  bases)  by  half 
their  common  perpendicular  (its  altitude). 

Ex.  290.  The  area  of  a  trapezoid  equals  the  product  of 
its  altitude  by  its  median  (the  sect  joining  the  bisection- 
points  of  the  non-parallel  sides). 

307.  To  find  the  altitudes  of  a  triangle  in  terms  of 
its  sides. 

B  Either  ^  A  or  ^  C  is  acute. 

Suppose  ^  C  acute, 

c2  =  a2  +  b2  —  2bj  (by  305). 

a2  +  ^2_c2 


h 


Fig.  119.  ,\  ; 

hb2=a2-j2=a 


26        * 
(a2  +  b2-c2)2     4a2b2-(a2  +  b2-c2)2 


4b2  4b2 

(2ab  +  a2  +  b2  -c2)(2ab  -a2  -b2  +  c2) 

4b2 
[(a  +  b)2-c2][c2-(a-b)2] 

4b2 
(a  +  b  +  c)(a  +  b-c)(c  +  a-b)(c-a  +  b) 
4b2 
Put  (a  +  b  +  c)=2S.     Then  a  +  b-c  =  2S-2C 
25     2(5  —c)2(s  —6)2(5  —  a) 


.\W:- 


4b'' 


;.hh  =l[s(s-a)(s-b)(s-c)]K 

308.  (Heron.)      To  find  the  area  of  a  triangle  in 

terms  of  its  sides. 

b    2  $ 

A=%bkb  =  -.-[s(s-a)(s-b)(s-c)]  ; 

,\  A  =[5(5  -a) (s  -b) (s-c)f. 


EQUIVALENCE. 


127 


Ex.291.  If  a2=b2+c2,  ;.^.A=r\^. 
lia2>b2+c2,  .'.  £A>r,t  £. 
Iia2<b2+c2,     :.  £A<r't£. 

Ex.  292.  Given  for  the  three  sides  of  a  triangle  numeri- 
cal expressions  in  terms  of  a  unit,  compute  the  three 
altitudes. 

Ex.  293.  The  sum  of  the  squares  of  two  sides  of  a  tri- 
angle is  equal  to  twice  the  square  of  half  the  third  side 
increased  by  twice  the  square  of  the  median  upon  that 
side.  [a2 +c2 —%b2  =  2mb2]. 

Ex.  294.  The  difference  of  the  squares  of  two  sides  of 
a  triangle  is  equal  to  twice  the  product  of  the  third  side 
by  the  sect  from  the  foot  of  that  side's  altitude  to  the 


foot  of  its  median. 


r.     a2-c2~l 


Ex.  295.  Given  numerical  expressions  for  the  sides  of 
a  triangle,  compute  the  medians. 

2mc  =  [2(a2+b2)-c2]i. 

309.  Theorem.  The  product  of  two  sides  of  a 
triangle  equals  the  product  of  two  sects  from  thai 
vertex  making  equal  angles  with  the  two  sides  ana 
extending,  one  to  the  base,  the  other  to  the  circle  cir- 
cumscribing the  triangle. 

Proof.     aCBD-aABE. 

310.  Corollary  I  to  309.  If 
BD  and  BE  coincide  they  bisect 
the  angle  B; 

:.AB-BC  =  BD-BE 
=  BD(BD+DE)=BD2  +  BD-DE 
=  BD2  +  CD-DA   (by  248). 

Therefore  the  square  of  a  bisecto? 
together  with  the  product  of  the  sects  Fig.  120. 

it  makes  on  a  side  equals  the  product  of  the  other  two 
sides. 


128  RATIONAL   GEOMETRY. 

311.  Corollary  II  to  309.  If  BD  be  an  altitude, 
BE  is  a  diameter,  for  then  ^.BAE  is  r't;  there- 
fore in  any  triangle  the  product  of  two  sides  equals  the 
product  of  the  diameter  of  the  circumscribed  circle  by 
the  altitude  upon  the  third  side. 

Ex.  296.  To  find  the  bisectors  of  the  angles  of  a  tri- 
angle, given  the  sides.  tc= j[abs{s  —  c)]l. 

Ex.  297.  To  find  the  radius  of  the  O  circumscribing  a  tri- 
angle. Rule:  Divide  the  product  of  the  three  sides  by 
four  times  the  area  of  the  triangle.  R=abc/4A. 

Ex.  298.  The  in-radius.  equals  area  over  half  sum  of 
sides.  [r=A/s]. 

Ex.  299.  The  side  of  an  equilateral  triangle  is 

Ex.  300.  The  radius  of  circle  circumscribing  triangle 
7,  15,  20,  is  12^.     The  in-radius  is  2. 

Ex.  301.  •  To  find  the  radius  of  an  escribed  circle.  Rule: 
Divide  the  area  of  the  triangle  by  the  difference  between 
half  the  sum  of  its  sides  and  the  tangent  side. 

[*-,«*/($ -a)J. 

Ex.  302.    A  =(rr1r2r3)i. 

Ex.303.   i/r1+i/r2+i/r3  =  i/r. 

Ex.  304.  The  sum  of  the  four  squares  on  the  four  sides 
of  any  quadrilateral  is  greater  than  the  sum  of  the  squares 
en  the  diagonals  by  four  times  the  square  on  the  sect 
joining  the  mid-points  of  the  diagonals. 

Ex.  305.  The  sum  of  the  squares  on  the  four  sides  of 
a  parallelogram  is  equal  to  the  sum  of  the  squares  on  the 
diagonals. 

Ex.  306.  The  product  of  the  external  segments  (sects), 
made  on  one  side  by  the  bisector  of  an  external  angle 
of  a  triangle  equals  the  square  of  the  bisector  together 
with  the  product  of  the  other  two  sides. 

Ex.  307.  Find  rv  r2,  r3,  when  a  =  7,  6  =  15,  c  =  20. 


EQUIVALENCE.  129 

The  Mensuration  of  the  Circle. 

312.  We  assume  that  with  every  arc  is  connected 
a  sect  such  that  if  an  arc  be  cut  into  two  arcs,  this 
sect  is  the  sum  of  their  sects;  moreover,  this  sect 
is  not  less  than  the  chord  of  the  arc,  nor,  if  the  arc 
be  minor,  is  it  greater  than  the  sum  of  the  sects  on 
the  tangents  from  the  extremities  of  the  arc  to 
their  intersection.  This  sect  we  call  the  length  of 
the  arc. 

313.  In  practical  science,  every  sect  is  expressed 
by  the  unit  sect  preceded  by  a  number. 

From  our  knowledge  of  the  number  and  the 
unit  sect  it  multiplies,  we  get  knowledge  of  the  sect 
to  be  expressed,  and  we  can  always  construct  this 
expression.  For  science,  the  unit  sect  is  the  centi- 
meter [cm],  which  is  the  hundredth  part  of  the  sect 
called  a  meter,  two  marked  points  on  a  special  bar 
of  platinum  at  Paris,  when  the  bar  is  at  the  tem- 
perature of  melting  ice. 

314.  If  an  angle  of  an  equilateral  triangle  be 
taken  at  the  center  of  a  circle,  the  chord  it  in- 
tercepts equals  the  radius.  Therefore  the  length 
of  a  semicircle  is  not  less  than  three  times  its 
radius. 

It  is  in  fact  greater,  since  joining  a  point  on 
the  arc  of  one  of  these  chords  to  its  extremities 
gives  a  pair  of  chords  together  greater  than  the 
radius. 

Again,  taking  any  diameter,  then  the  diameter 
perpendicular  to  this,   then  perpendiculars  at  the 


130  RATIONAL   GEOMETRY. 

four  extremities  of  these,  we  have  a  square  of  tan- 
gents equal  to  Sr. 

Therefore  the  length  of  a  semicircle  is  not  greater 
than  four  times  its  radius.  It  is  in  fact  less,  as  is 
seen  by  drawing  a  tangent  at  any  fifth  point  of  the 
circle.  The  number  prefixed  to  the  radius  in  the 
expression  for  the  length  of  the  semicircle  is  desig- 
nated by  the  symbol  n. 

The  length  of  any  circle  is  27tr. 

So  the  lengths  of  circles  are  proportional  to  their 
radii. 

Historical  Note  on  n. 

315.  We  have  proved  that  n  is  greater  than  3 
and  less  than  4,  but  the  Talmud  says:  "What  is 
three  handbreadths  around  is  one  handbreadth 
through,"  and  our  Bible  also  gives  this  value  3. 
[I  Kings  vii.  23;   II  Chronicles  iv.  2.] 

Ahmes  (about  1700  b.c.)  gave  [4/3]*  =  3- 16. 
Archimedes  placed  it  between  3^ r  and  $\.  Ptolemy 
used  3TW. 

The  Hindoo  Aryabhatta  (b.  476)  gave  3-1416; 
the  Arab  Alkhovarizmi  (nourished  813-833)  gave 
3-1416;  Adriaan  Anthoniszoon,  father  of  Adriaan 
Metius  [in  1585]  gave  355/113=3-1415929;  Ludolf 
van  Ceulen  [1540-16 10]  gave  the  equivalent  of  over 
30  decimal  places 

[71  =  3-141592653589793238462643383279+] 
(the  decimal  fraction  was  not  yet  invented),  and 
wished  it  cut  on  his  tomb  at  Leyden.     Vega  gave 
140  cjecimal  places;   Dase,  200;   Richter,  500. 


EQUIVALENCE. 


I31 


In  1873  Wm.  Shanks  gave  707  places  of  decimals. 

The  symbol  it  is  first  used  for  this  number  in 
Jones's  ''Synopsis  Palmariorum  Matheseos,"  Lon- 
don, 1706. 

In  1770  Lambert  proved  tz  irrational,  that  is, 
inexpressible  as.  a  fraction.  In  1882  Lindemann 
proved  x  transcendent,  that  is,  not  a  root  of  any 
algebraic  equation  with  rational  coefficients,  and 
hence  geometrically  inconstructable. 

316.  Kochansky  (1685)  gave  the  following  simple 
construction  for  the  length  of  the  semicircle: 


Fig.   121. 


At  the  end-point  A  of  the  diameter  BA  draw 
the  tangent  to  the  circle  OC(CA).  Take  ^fACE  = 
half  the  angle  of  an  equilateral  triangle  =  J  r't  ^ . 
On  the  tangent,  take  EF  =  y. 

Then  BF  is  with  great  exactitude  the  length  of 
the  semicircle.    In  fact  BF  =  r[i^  —  2(3)*]* -s3*  141  $r. 

317.  Definition.  The  circle  with  the  unit  sect 
for  radius  is  called  the  unit  circle. 

318.  Definition.  The  length  of  the  arc  of  unit 
circle  intercepted  by  an  angle  with  vertex  at  center 
is  called  the  size  of  the  angle. 


I32  RATIONAL    GEOMETRY. 

319.  Definition.  The  angle  whose  size  is  the 
unit  sect  is  called  a  radian. 

320.  Theorem.  A  radian  intercepts  on  any  circle 
an  arc  whose  length  is  that  circle's  radius. 

321.  Corollary  to  320. 

If  u  denote  the  number  of  radians  in  an  angle 
and  I  the  length  of  the  arc  it  intercepts  on  circle 
of  radius  r,  then 

u=l/r. 

322.  Definition.  An  arc  with  the  radii  to  its 
end-points  is  called  a  sector. 

323.  Definition.  The  area  of  a  sector  is  the 
product  of  the  length  of  its  arc  by  half  the  radius. 

324.  Corollary  to  323  and  314. 
The  area  of  any  circle,  o  =r27r. 

325.  Corollary  to  324. 

The  areas  of  circles  are  proportional  to  the  areas 
of  squares  on  their  radii. 

Ex.  308.  The  areas  of  similar  polygons  are  as  the  squares 
of  corresponding  sides. 

Ex.  309.   Find  the  length  of  the  circle  when  r  =  14  units. 

Ex.  310.  Find  the  diameter  of  a  wheel  which  in  a  street 
19,635  meters  long  makes  3125  revolutions. 

Ex.  311.  Find  the  length  and  area  of  a  circle  when 
r  =  7. 

Ex.  312.  If  we  call  one-ninetieth  of  a  quadrant  a  degree 
of  arc  and  its  angle  at  the  center  a  degree  of  angle,  find  the 
size  of  this  ^  (size  of  ^i°). 

Ex.  313.   How  many  degrees  in  a  radian? 

Ex.  314.  The  angles  of  a  triangle  are  as  1:2:3.  Find 
the  size  of  each.     Find  the  number  of  degrees  in  each. 

Ex.315.  The  angles  of  a  quadrilateral  are  as  2:3:4:  7. 
Find  each  in  degrees  and  radians. 


EQUIVALENCE.  133 

Ex.  316.   In  what  regular  polygon  is  every  angle  168^°? 

n 
Ex.  317.   If  a  r't^   be  divided  into  —    congruent   parts, 

how  many  of  them  would  a  radian  contain? 

Ex.  318.  Find  the  length  of  the  arc  pertaining  to  a 
central  angle  of  7 8°  when  r=i-5  meters. 

Ex.  319.  Find  an  arc  of  1120  which  is  4  meters  longer 
than  its  radius. 

Ex.  320.  Calling  7:=%?-,  find  r  when  640  are  70-4  meters. 

Ex.  321.  Find  the  inscribed  angle  cutting  out  one-tenth 
of  the  circle. 

Ex.  322.  An  angle  made  by  two  tangents  is  the  differ- 
ence between  1800  and  the  smaller  intercepted  arc.  Make 
this  statement  exact. 

Ex.  323.  Find  the  size  of  half  a  right  angle. 

Ex.  324.  Find  the  size  of  300;    450;    6o°. 

Ex.  325.   How  many  radians  in  r°?  in  2400? 

Ex.  326.  Express  the  size  of  seven-sixteenths  of  a  right 
angle. 

Ex.  327.  How  many  radians  in  the  angle  made  by  the 
hands  of  a  watch  at  5:  15  o'clock?  at  quarter  to  8?  at 
3:  30?  at  6:  05?  » 

Ex.  328.  The  length  of  half  a  quadrant  in  one  circle 
equals  that  of  two-thirds  of  a  quadrant  in  another.  Find 
how  many  radians  would  be  subtended  at  the  center  of 
the  first  by  an  arc  of  it  equal  in  length  to  the  radius  of 
the  second. 

Ex.  329.  Find  the  number  of  degrees  in  an  angle  whose 
size  is  i;   is  $;   is  f ;   is  |[;   is  %n. 

2  •  «t 
Ex.  330.  The  size  of   the   sum   of   two   angles   is   — ** 

and  their  difference  is  170;   find  the  angles. 

Ex.  331.  How  many  times  is  the  angle  of  an  isosceles 
triangle  which  is  half  each  angle  at  the  base  contained 
in  a  radian? 

Ex.  332.  Two  wheels  with  fixed  centers  roll  upon  each 
other,  and  the  size  of  the  angle  through  which  one  turns 
gives   the   number   of   degrees    through   which   the    other 


134  RATIONAL    GEOMETRY. 

turns  in  the  same  time.  In  what  proportion  are  the  radii 
of  the  wheels? 

Ex.  $3$.  The  length  of  an  arc  of  6o°  is  36I;  find  the 
radius. 

Ex.  334.  Find  the  circle  where  ^30°  is  subtended 
by  arc  4  meters  long. 

Ex.  335.   If  0  be  area  of  circle,  prove 

i/tOA^^  +  i/tO/^r^  +  i/tO^Wl^iAG/Wli 

Ex.  336.  The  perimeters  of  an  equilateral  triangle,  a 
square,  and  a  circle  are  each  of  them  12  meters.  Find 
the  area  of  each  of  these  figures  to  the  nearest  hundredth. 

Ex.  337.  An  equilateral  triangle  and  a  regular  hexagon 
have  the  same  perimeter;  show  that  the  areas  of  their 
inscribed  circles  are  as  4  to  9. 

Ex.  338.  Find  the  number  of  degrees  in  the  arc  of  a 
sector  whose  area  equals  the  square  of  its  radius. 

Ex.  339.  Find  area  of  sector  whose  radius  equals  25  and 
the  size  of  whose  angle  is  f, 

Ex.  340.  The  length  of  the  arc  of  a  sector  is  16  meters, 
the  angle  is  £  of  a  r't  ^.     Find  area  of  sector. 

Ex.  341.  If  2As  have  a  common  base,  their  areas  are 
as  the  segments  into  which  the  join  of  the  vertices  is  di- 
vided by  the  common  base. 

Ex.  342.  The  area  of  a  circum-polygon  is  half  perimeter 
by  in-radius  [%pr]. 

Ex.  343.  The  area  of  a  rhombus  is  half  the  product  of 
its  diagonals. 

Ex.  344.  If  we  magnify  a  quad'  until  a  diagonal  is 
tripled,  what  of  its  area? 

Ex.  345.  If  the  sum  of  the  squares  on  the  three  sides 
of  a  A  =  8  times  the  square  of  a  median  the  A  is  r't-angled. 

Ex.  346.  Lengthening  through  A  the  side  b  of  a  a 
by  c  and  c  by  b,  they  become  diagonals  of  a  symtra  whose 
area  is  to  that  of  the  A  as  (b  +  c) 2  to  be. 

Ex.  347.  If  upon  the  three  sides  of  a  r't  A  as  corre- 
sponding sides  similar  polygons  are  constructed  that  on 
the  hypothenuse  =the  sum  of  those  on  the  Is. 


EQUIVALENCE.  135 

Ex.  348.  The  area  of  any  r'tA  =the  sum  of  the  areas  of 
the  two  limes  or  crescent-shaped  figures  made  by  de- 
scribing semi- os  outwardly  on  the  jls  and  a  semi-o  on 
the  hypothenuse  through  the  vertex  of  the  r't  if  [called 
the  lunes  of  Hippocrates  of  Chios  (about  450  b.c.)]. 

Ex.  349.  (Pappus.)  Any  two  ||g'ms  on  two  sides  of  a  A 
are  together  —to  a  ||g'm  on  the  third  side  whose  consecutive 
side  is  =  and  ||  to  the  sect  joining  the  common  vertex  of 
the  other  ||g'ms  to  the  intersection  of  their  sides  ||  to  those 
of  the  A  (produced) . 

Ex.  350.  If  all  the  sides  of  a  quad'  are  unequal,  it  is 
impossible  to  divide  it  into  =  a  s  by  straights  from  a  point 
within  to  its  vertices. 

Ex.  351.  The  joins  of  the  centroid  and  vertices  of  a  a 
trisect  it. 

Ex.  352.  Make  a  symtra  triple  a  given  symtra. 

Ex.  353.  On  each  side  of  a  quad'  describe  a  sq'  out- 
wardly. Of  the  four  as  made  by  joining  their  neighbor- 
ing corners,  two  opposite  =the  other  two  and  =  the  quad'. 

Ex.  354.  If  from  an  ^  a  we  cut  two  =  as,  one  I-  ,  the 
sq'  of  one  of  the  =  sides  of  the  I-  a  equals  the  product  of 
the  sides  of  the  other  A  on  the  arms  of  the  7f  a. 

Ex.  355.  If  any  point  within  a  |!g'm  be  joined  to  the 
four  vertices,  one  pair  of  as  with  ||  bases   =  the  other. 

Ex.  356.  One  median  of  a  trapezoid  cuts  it  into  =parts. 

Ex.  357.  Transform  a  given  a  into  an  —  +  A. 

Ex.  358.  Transform  a  given  •(•  a  into  a  regular  a. 

Ex.  359.  Construct  a  polygon  ~  to  two  given  ~  poly- 
gons and  =  to  their  sum. 

Ex.  360.  If  a  vertex  of  a  a  moves  on  a  1  to  the  oppo- 
site side,  the  difference  of  the  squares  of  the  other  sides 
is  constant. 

Ex.  361.  The  ^  bi's  of  a  rectangle  make  a  sq',  which 
is  half  the  sq'  on  the  difference  of  the  sides  of  the  rectangle. 

Ex.  362.  The  bisectors  of  the  exterior  ^  s  of  a  rectangle 
make  a  sq'  which  is  half  the  sq'  of  the  sum  of  the  sides  of 
the  rectangle. 

Ex.  363.  The  sum  of  the  squares  made  by  the  bisectors 


I36  RATIONAL   GEOMETRY. 

of  the  interior  and  exterior  j£  s  of  a  rectangle  equals  the 
sq'  of  its  diagonal;  their  difference  is  double  the  rect- 
angle. 

Ex.  364.  If  on  the  hypothenuse  we  lay  off  from  each 
end  its  consecutive  side,  the  sq'  of  the  mid  sect  is  double 
the  product  of  the  others. 

Ex.365.   In   aABC,  BD-a-BF-c. 

Ex.  366.  In  a  trapezoid,  the  sum  of  the  sq's  on  the 
diagonals  equals  the  sum  of  the  sq's  on  the  non-||  sides 
plus  twice  the  product  of  the  ||  sides. 

Ex.  367.  Prove  r1+r2+r3  =  r+4i?. 

Ex.  368.  To  bisect  a  a  by  a  st'  through  a  given  point 
in  a  side;  by  a  st'  ||  to  a  side;    _L  to  a  side. 

Ex.  369.  Trisect  a  -I-  a  by  ||s. 

Ex.  370.  A  quad'  equals  a  a  with  its  diagonals  and 
their   ^   as  sides  and  included   ^  . 

Ex.  371.  The  areas  of  as  inscribed  in  a  ©  are  as  the 
products  of  their  sides. 

Ex.  372.  Construct  an  equilateral  a,  given  the  altitude. 

Ex.  373.    a  from  ^s  and  area. 

Ex.  374.  Triple  the  squares  of  the  sides  of  a  A  is  quad- 
ruple the  sq's  of  the  medians. 

Ex.  375.  Any  quad'  is  divided  by  its  diagonals  into 
four  as  whose  areas  form  a  proportion. 

Ex.376.  AH-HD=BH-HE. 

Ex.  377.  The  area  of  a   -I-   r't  A  is  $c2. 

Ex.  378.  Construct  i  A  =given   a  with  same  b  and  hb. 

Ex.  379.  Bisect  any  quad'  by  a  st'  from  any  vertex;  from 
any  point  in  a  side. 

Ex.  380.  Any  st'  through  the  bisection-point  of  a  diag- 
onal bisects  the  ||gm. 

Ex.  381.  3(a2  +  62  +  c2)  =4(ma2  +  m2  +  wc2). 

Ex.  382.  Upon  any  st'  the  sum  of  the  ±s  from  the 
vertices  of  a  a  is  thrice  the  1  from  its  centroid. 

Ex.383.   In  r't  A,  $c2 ■=4(wo,  +  Wt2). 

Ex.  384.  Trisect  a  quad'. 

Ex.  385.  Find  a  =aABC,  but  with  sides  m,  n;  with 
side  m  and  adjoining  ^  d;   and  opposite  ^  8. 


EQUIVALENCE.  137 

Ex.386.  Find  -I-  A  =  A  ABC,  but  with  base  m\  with 
side  m. 

Ex.  387.  Find  a  =  given  polygon. 

Ex.  388.  From  any  point  in  an  equilateral  a  the  three  ±s 
on  the  sides  together  =the  altitude. 

Ex.  389.  Sects  from  the  bisection-point  of  a  non-||  side 
of  a  trapezoid  to  oppsite  vertices  bisect  it. 

Ex.  390.  If  the  products  of  the  segments  of  two  inter- 
secting sects  are  =,  their  ends  are  concyclic. 

Ex.391.  Area  of  r't  A  =  product  of  the  segments  of 
the  hypothenuse  made  by  _|_  from  /. 

Ex.  392.  In  r't  a,  areas  of  as  made  by  he  are  propor- 
tional to  areas  of  their  in-os. 

Ex.393.   1  /ha  +  1  /hb  +  1  /he  =  1  /r. 

Ex.394.  hahbhc=(a  +  b  +  c)3r3/abc. 

Ex.  395.  If  ha,  hb',  he  be  the  perpendiculars  from  any 
point  within  a  a,  upon  the  sides,  prove  haf /ha-\-hbr /lib 
+  hc,/he  =  i. 

Ex.396.  r=\AI-BI-CI(a  +  b  +  c)/abc. 

Ex.  397.  abc  =a{Aiy  +  b{Biy  +  c(CI)\ 

Ex.  398.  (AI)2+(BI)2+(CI)2=  ab  +  ac  +  be-  6abc/(a  + 
b+c). 

Ex.  399.  R  +  r  ■=  _Ls  from  0  on  sides. 

Ex.  400.   In  -I-    a,  if  b  =hb,  then  fb  =R. 

Ex.  401.  R  =2R  of  aDEF. 

Ex.402.  Area  of  a  /,,  I2,  I3=abc/2r. 

Ex.  403.  If  qa,  qb,  qc  be  the  sides  of  the  3  sq's  inscribed 
in  a  a,  then  1  /qa  ■=  1  Jha  +  1  /a ;  1  /qb  =  1  /hb  +  1  /b ;  i/qc 
=  i/he+i/c 

Ex.  404.  1  /r  =  1  /ha  +i/hb+i/hr;  1  /r,  =  -  1  /ha  +  1  /hb 
+  i/hc;    1  frt  =  1  /ha  —  i/hb+i/hc;    1  /r3  -  1  /ha  +  1  /hb  —  1  /he. 

Ex.405.  2/ha  =  i/r -i/rl=i/r2+i/r3;  2 /hb  =  i/r  —  1  /r2 
=  1  /r3+  1  /r, ;  2/hc  =  i/r-i /r3  =  1  /r,  +  1  /r2. 

Ex.  406.  Ju/ 2  =rrj{r,  -r)  =r2r3/(rz  +  r3). 

Ex.407.  R2=(IOy+2rR  =  (IlOy-2rlR. 

Ex.  408.  Find  the  segments  of  b  made  by  tb. 

Ex.  409.  The  base  of  a  a  is  32  feet  and  its  height  20  feet; 


I38  RATIONAL    GEOMETRY. 

what  is  the  area  of  the  a  formed  by  drawing  a  st'  ||  b 
5  feet  from  B  ?  Where  must  a  st'  ||  b  be  drawn  so  as  to 
divide  the  a  into  2  parts  of    m  area? 

Ex.  410.  Upon  each  side  a  of  a  sq'  as  diameter  semi- 
circles are  described  within  the  sq',  forming  4  leaves; 
find  the  area  of  a  leaf. 


CHAPTER  XI. 

GEOMETRY  OF  PLANES. 

326.  Theorem.     Two  parallels  determine  a  plane. 
Proof.     By   definition   they  are   coplanar.     Any 

plane  on  these  parallels  would  be  on  three  non- 
cost  raight  points  of  this  given  plane,  hence  (by  I  4) 
identical  with  it. 

327.  Corollary  to  326.  If  a  plane  contains  one 
of  two  parallels  and  any  point  of  the  other,  it  con- 
tains both  parallels. 

328.  Theorem.  Three  planes  which  do  not  con- 
tain the  same  straight  cannot  have  more  than  one 
point  in  common. 

Proof.  If  they  had  two  points  in  common 
(by  I  5)  the  straight  determined  by  those  two 
points  would  be  in  each. 

329.  Corollary  to  328.  If  three  planes  not  con- 
taining the  same  straight  intersect  in  pairs,  the 
three  straights  of  intersection  [common  sections,  or 
meets']  are  either  copunctal  or  parallel  in  pairs. 

330.  Corollary  to  329.  If  a  plane  on  one  of  two 
parallels  meet  a  plane  on  the  other  [neither  that 
of  the  || s],  the  meet  is  parallel  to  each  of  the  two 
parallels. 

For  (by  9)  the  three  planes  can  have  no  point 

x39 


140 


RATIONAL    GEOMETRY. 


in  common,  since  a  point  common  to  the  three  would 
be  common  to  the  two  parallels. 

Ex.  411.  If  a  st'  cross  three  copunctal  st's,  the  four 
are  copunctal  or  coplanar. 

Ex.  412.  If  each  of  three  st's  crosses  the  others,  the 
three  are  coplanar  or  copunctal. 

Ex  413.  The  meet  of  planes  determined  by  two  pairs 
of  st's  on  A  is  on  A. 

Ex.  414.  If  A  is  on  a  and  a,  it  is  on  the  intersections 
of  a  with  planes  on  a. 

331.  Problem.  Through  a  given  point  A  of  a  given 
plane  a  to  pass  straights  a  and  b  in  a. 


Fig.   122. 

Solution.  There  are  (by  I  7)  in  the  plane  at 
least  three  non-costraight  points,  A,  B,  C.  But 
(by  I  5)  A  and  B  determine  a  straight  in  the  plane  a. 
So  do  A  and  C. 

332.  Problem.     To  put  two  planes  on  the  straight  a. 


Fig.  n?3. 

Solution.     On  a  (by  I  2)  are  at  least  two  points, 
A  and  B.     There  are  (by  I  7)  at  least  four  non- 


GEOMETRY   OF  PLANES. 


141 


costraight  non-coplanar  points,  A,  B,  C,  D.  There- 
fore A,  B,  and  C  are  not  costraight,  otherwise 
(by  11)  A,  B,  C,  D  would  be  coplanar.  Therefore 
(by  I  3)  A,  B,  C  determine  a  plane  which  (by  I  5) 
contains  a. 

Just  so  A,  B,  D  determine  a  plane  on  a. 

333.  Theorem.  If  a  straight  be  perpendicular 
to  each  of  two  intersecting  straights,  it  will  be  per- 
pendicular to  every  other  straight  in  their  plane  and 
on  their  point  of  intersection. 

Hypothesis.  Let  BP  be  ±  to  BA.  and  BC. 
Let  BD  be  in  plane  ABC. 

Conclusion.     BP±BD. 


Fig.   124. 

Proof.  Take  A  and  C  on  the  sides  of  the  angle 
at  B  in  which  BD  lies.  Let  D  be  the  point  where 
AC  crosses  BD.  Take  BP' =BP.  Then  (by  43), 
aPBA  =  aP'BA,  and   aPBC=aP'BC. 


142 


RATIONAL    GEOMETRY. 


:.  PA  =  P'A  and  PC=P'C;   .  ■  .  (by  58)  aPAC^ 
aP'AC. 
/.  4PCD=4P'CD\    /.  (by  43)  aPCD^aP'CD. 
:.PD=P'D\    .'.(by  58)   aPBD=  aP'DB. 
/.  ^PBD^^P'BD;   ;.BP\_BD. 

334.  Definition.  A  straight  is  said  to  be  per- 
pendicular to  a  plane  when  it  is  perpendicular  to 
every  straight  in  that  plane  which  passes  through 
its  foot, — that  is,  the  point  it  has  in  common  with 
the  plane,  called  also  their  pass. 

Then  also  the  plane  is  said  to  be  perpendicular 
to  the  straight. 

335.  Definition.  A  straight  is  said  to  be  parallel  to  a 
plane  when  it  has  no  point  in  common  with  the  plane. 

Then  also  the  plane  is  said  to  be  parallel  to  the 
straight. 

336.  Definition.  A  straight  neither  on  the  plane, 
nor  parallel  nor  perpendicular  to  the  plane,  is  said 
to  be  oblique  to  the  plane.  A  sect  from  a  point  to  a 
plane,  if  it  be  not  perpendicular,  is  called  an  oblique. 


Fig.  125. 
337.  Problem.     To  construct  a  plane  perpendicular 
to  a  given  straight  a  at  a  given  point  A. 


GEOMETRY  OF  PLANES. 


143 


Solution.  Put  (by  332)  through  a  two  planes. 
In  each  of  them  at  A  (by  161)  erect  a  perpendicular 
to  a.  The  plane  of  these  perpendiculars  is  (by  333 
and  334)  perpendicular  to  a  at  A. 

338.  Problem.  To  construct  a  plane  perpendicular 
to  a  given  straight  a  through  a  given  point  P  not  on  a. 

Solution.     By  160,  drop  PA±a. 
By  332  and  161,  erect  another  perpendicular  A B 
to  a  at  A.     Then  (by  333  and  334)  plane  PAB\_a. 

339.  Problem.  To  erect  a  perpendicular  to  a  given 
plane  y  at  a  given  point  A . 

Solution.  Take  (by  331)  through  A  two  straights, 
a,  b,  in  the  plane  y.  Find  (by  337)  a  plane  a  which 
at  A  is  _L  to  a ;  also  a  plane  ^  which  at  A  is  _L  b. 


Fig.   126. 


These  two  planes  (by  329)  intersect  in  a  straight 
c  through  A. 

Since  c  is  in  a,  .*.  (by  334)  c  is  _L  a. 
Since  c  is  in  /?,  ,\  c  is  JL  b. 
.'•  (by  333  and  334)  c±T- 


44 


RATIONAL    GEOMETRY. 


340.  Corollary  to  339.  Through  a  given  point  in 
a  plane  there  is  only  one  perpendicular  to  that 
plane. 


Fig. 


Else  each  would  be  JL  to  BC,  the  meet  of  their 
plane  /?  with  the  given  plane  a. 

341.  Problem.  To  drop  a  perpendicular  to  a  given 
plane  a  from  a  given  point  P. 

Construction.     In  a  take  any  straight  a. 


From  P  (by   160)  drop  PA±a.     In  a  (by  161) 
erect  b±a.     From  P  drop  PB±b.     Then  PB±a. 


GEOMETRY  OF  PLANES.  M5 

Proof.  Produce  PB  through  B,  taking  BP'  =  BP. 
Then  (by  43)  aPBA^aP'BA;  /.PA=P'A. 
Further,  since  a±AP  and  a_\_AB,  .'.  also  (by  333) 
a±AP'.  Thus  if  M  be  a  second  point  on  a, 
,\r't   ^PM=^P'M. 

.-.  (by  43)    aPM^aP'M;    ;.PM=P'M. 

/.(by  58)   aPBM=aP'BM\    ;.^PBM  is  r't. 

But  by  construction  ifPBA  is  r't,  .".  (by  333 
and  334)  PB±a. 

342.  Corollary  to  341.  From  a  point  P  without 
a  plane  a,  there  is  only  one  perpendicular  to  the 
plane  a. 

Take  (by  341)  PBl_a.  Then  if  A  be  any  other 
point  of  a,  the  r't  aPBA  has  (by  79)  the  4PAB 
acute. 

343.  Corollary  to  342.  From  a  point  to  a  plane, 
the  perpendicular  is  less  than  any  oblique.  Equal 
obliques  meet  the  plane  in  a  circle,  whose  center 
is  the  foot  of  the  perpendicular.  If  through  the 
center  of  a  circle  a  perpendicular  to  its  plane  be 
taken,  then  sects  from  a  point  of  this  perpendicular 
to  the  circle  are  equal. 

344.  Theorem.  If  a  straight  is  perpendicular  to 
each  of  three  straights  copunctal  with  it,  the  three 
are  coplanar. 

Hypothesis.     PB±_BA,  BD,  BC. 

Conclusion.     BC  in  plane  BDA [a]. 

Proof.  Let  plane  PBC[p]  meet  plane  a  in  BC . 
Then  (by  333)  PB±BC.  By  hypothesis  PB1.BC. 
But  (by  52)  in  /?  is  only  one  perpendicular  to  PB 
at  B.     .'.  BC  is  identical  with  BC. 

345.  Corollary  to   344.     Through  a  given   point 


146 


RATIONAL    GEOMETRY. 


in  a  straight  there  is  only  one  plane  perpendicular 
to  that  straight. 


Fig.  129. 

346.  Theorem.  All  points,  A,  which  with  two 
fixed  points  B,  C  given  equal  sects,  AB=AC,  are 
in  the  plane  a  bisecting  at  right  angles  the  sect 

BC,     and     inversely    every 
point    Af    in    the    plane    a 
bisecting    at    right    angles 
the    sect    BC    gives    A'B  = 
A'C. 

Proof.  For  the  straight 
from  A  to  the  bisection 
point  D  of  BC  makes  a  A  DB 
=  aADC  and  .'.(by  344) 
is  in  a. 

Inversely    every    straight 
A'D  is  ±BC  and   .*.  makes 
aA'DB=aA'DC,  and  ;.A'B=A'C. 

347.  Corollary  to  338.  Through  a  given  point 
without  a  given  straight  there  is  only  one 
plane  perpendicular  to  that  straight. 

Take  (by  33%)   a  through  P  and  1  to  a  at  B. 


Fig.  130. 


GEOMETRY  OF  PLANES. 


147 


Then  if  A  is  any  other  point  of  a,  the  r't  aPBA 

has    (by    79)    the    ifPAB 

acute.     So  plane  p  through 

P  and  1  to    a    could  not 

pass   through  A,   and  so, 

passing  through  B,  it  is  (by 

345)  identical  with  a. 

348.  Theorem.  If  PB±- 
BAC  and  BA±AC,  then 
PA±AC. 

Proof.     Make   AC=BP. 
.-.(by        43)         aCAB  = 
aPBA.  .-.CB=PA,    .'.   (by   58)    aCBP 
.-.  tCBP^tPAC. 


aPAC, 


Fig.   132. 

But  by  hypothesis   %-CBP  is  right. 

Ex.  415.   If  PBLBAC  and  PA±AC,  then  BAlAC. 

Ex.  416.  If  PBLBAC  and  BA±AC,  all  is  to  AC  from 
points  in  PB  go  to  A. 

Ex.417.  If  PHLABC  (//  is  orthocenter),  then  PA± 
to  AK  H  J5C. 

349.  Theorem.  Two  perpendiculars  to  a  plane 
are  coplanar. 


148  RATIONAL    GEOMETRY. 

Hypothesis.     PB,  P'A\_a  at  B,  A. 
Conclusion.     P'  is  in  the  plane  ABP. 
Proof.     In  a  erect  AC±AB. 

.'.    (by    348)     AC±AP.       But     by     hypothesis 
AC\AP'. 

.'.  (by  344)  Ay  P',  P,  B  are  coplanar. 


p' 


/ 


Fig.   133. 

350.  Corollary  to  349  and  342.  Two  perpen- 
diculars to  a  plane  are  parallel. 

351.  Inverse  of  350.  If  the  first  of  two  parallels 
is  perpendicular  to  a  plane,  the  second  is  also  per- 
pendicular to  that  plane. 

For  the  perpendicular  erected  to  the  plane  from 
the  foot  of  the  second  is  (by  350)  parallel  to  the 
first  and  so  (by  IV)  identical  with  the  second. 

352.  Theorem.  If  one  plane  be  perpendicular 
to  one  of  two  intersecting  straights,  and  a  second 
plane  perpendicular  to  the  second,  they  meet  and 
their  meet  is  perpendicular  to  the  plane  of  the  two 
straights. 

Hypothesis.  Let  a  be  ±CA  at  A  and  /?  be 
±CB  at  B. 

Proof.  The  meet  AD  of  a  with  plane  ACB  is 
(by  334)  ±AC,  and   likewise  BD±BC;    .'.  (by  77). 


GEOMETRY  OF  PLANES. 


149 


AD  meets  BD.  Thus  the  planes  a  and  /?,  having 
D  in  common,  meet  in  DP,  and  (by  351)  their 
meet  PD  is  1  to  a  straight  through  D  ||  to  AC, 
and  also  X  to  a  straight  through  D  ||  to  BC. 


Fig.   134. 

353.  Theorem.  Two  straights,  each  parallel  to 
the  same  straight,  are  parallel  to  one  another,  even 
though  the  three  be  not  co planar. 

For  a  plane  J_  to  the  third  will  (by  351)  be  J_ 
to  each  of  the  others;    .'.  (by  350)  they  are  ||. 

Ex.  418.  Are  st's  ||  to  the  same  plane  ||?  Are  planes  || 
to  the  same  st'  ||? 

Ex.  419.  A  plane  ||  to  the  meet  of  two  planes  meets 
them  in  ||s. 

354.  Definition.  The  projection  of  a  point  upon 
a  plane  is  the  foot  of  the  perpendicular  from  the 
point  to  the  plane. 

The  projection  of  a  straight  upon  a  plane  is  the 
assemblage  of  the  projections  of  all  points  of  the 
straight. 

355.  Theorem.  The  projection  of  a  straight  on 
a  plane  is  the  straight  through  the  projections  of  any 
two  of  its  points. 


*$o 


RATIONAL    GEO  ME  I RY. 


Given  A',  P\  B',  the  projections  of  A,  P,  B, 
points  of  the  straight  AB,  on  the  plane  a. 

To  prove  P*  in  the  straight  A'B'. 

Proof.  A}  A',  B,  Br  are  (by  349)  coplanar. 
P  is  (by  I  5)  in  this  same  plane,  .'.  (by  350  and 
327)  so  is  PP'  \   :.  (by  9)  A',  P',  B'  are  costraight. 


Fig.  135. 

356.  Corollary  to  355.  A  straight  and  its  pro- 
jection on  a  given  plane  are  coplanar.  If  a  straight 
intersects  a  plane,  its  projection  passes  through  the 
point  of  intersection.  A  straight  parallel  to  a  plane 
is  parallel  to  its  projection  on  that  plane. 

357.  Theorem.     A    straight   makes    with    its    own 


Fig.  136. 

projection  upon  a  plane  a  less  angle  than  with  any 
other  straight  in  the  plane. 


GEOMETRY  OF  PLANES.  151 

Hypothesis.  Let  A'  and  BA'  be  the  projection 
of  A  and  BA  on  a,  and  BC  any  other  straight  in  a, 
through  B. 

Conclusion.      4  ABA'  <  *  ABC. 

Proof.  Take£C  =  £A'.  Then  A  A'  <  AC.  (The 
perpendicular  is  the  least  sect  from  a  point  to  a 
straight.) 

/.  4  ABA' <4  ABC. 

(In  two  A's,  if  a=a',  b=b'y  c<c',  then^C<  4  C.) 

358.  Definition.  The  angle  between  a  straight 
and  its  projection  on  a  plane  is  called  the  inclina- 
tion of  the  straight  to  the  plane. 

Ex.  420.  One  of  three  copunctal  st's  makes  =  £s  with 
the  others  if  its  projection  on  their  plane  bisects  their  •£ . 

Ex.  421.  An  oblique  makes  with  some  st'  in  the  plane 
through  its  foot  any  given  ^  <  the  supplement  of  its 
inclination  and  >  its  inclination. 

Ex.  422.  Equal  obliques  from  a  point  to  a  plane  are 
equally  inclined  to  it. 

359.  Definition.  Parallel  planes  are  such  as 
nowhere  meet. 

360.  Theorem.  Planes  perpendicular  to  the  same 
straight  are  parallel. 

Proof.  They  cannot  (by  345  and  347)  have  a 
point  in  common. 

Ex.  423.  A  st'  and  a  plane  JL  to  the  same  st'  are  ||. 

361.  Theorem.  Every  plane  through  only  one  of 
two  parallels  is  parallel  to  the  other. 

Given  AB\\  CD  in  a,  and  /?  another  plane  through 
AB. 

To  prove  CD  ||  /?. 


*5 


RATIONAL   GEOMETRY. 


Since  AB  is  in  a  and  in  /?,  it  contains  (by  9) 
every  point  common  to  the  two  planes. 


Fig.   r37. 

But  CD  is  wholly  in  a.  So  to  meet  fi  it  must 
have  a  point  in  common  with  a  and  /?,  that  is,  it 
must  meet  A B.     But  by  hypothesis  AB  \\  CD. 

Ex.  424.  Through  a  given  point  to  draw  a  st'  ||  to  two 
given  planes. 

Ex.  425.  If  a  ||  a,  and  b  the  meet  of  a  with  /?,  /?  on  a 
then  a  \\  b. 

Ex.  426.  Through  A  determine  a  to  cut  b  and  c. 

362.  Problem.  Through  either  of  two  straights  not 
co planar  to  pass  a  plane  parallel  to  the  other. 


Fig.  138. 
If   AB   and   CD   are   the   given   straights,    take 
CF  ||  AB.     Then  (by  361)  DCF  \\  AB. 


GEOMETRY  OF  PLANES 


*53 


Determination.  There  is  only  one  such  plane. 
For,  through  DC  any  plane  ||  A B  meets  plane  ABC 
in  the  parallel  to  AB  through  C,  .'.is  identical 
with  CDF. 

Ex.  427.  Through  a  point  without  a  plane  pass  any 
number  of  st's  ||  to  that  plane. 

Ex.  428.  Through  a  point  without  a  st'  pass  any  number 
of  planes  ||  to  that  st'. 

Ex.  429.   Planes  on  a  \\  a  meet  «  in  ||s. 

Ex.  430.   If  a  I|  a  and  a  ||  p,  then  a  ||  to  the  meet  a/?. 

363.  Problem.  Through  any  given  point  P  to 
pass  a  plane  parallel  to   any  two   given   straights, 


Fig.  139. 

a,  b.  [The  plane  determined  by  the  parallel  to  a 
through  P,  and  the  parallel  to  b  through  P.]  [There 
is  only  one  such  plane.] 

Ex.  431.  Through  two  non-coplanar  straights  one  and 
only  one  pair  of  ||  planes  can  be  passed. 

364.  Theorem.  The  intersections  of  two  parallel 
planes  with  a  third  plane  are  parallel. 

Proof.  They  cannot  meet,  being  in  two  parallel 
planes;  yet  they  are  coplanar,  being  in  the  third 
plane. 


J54 


RATIONAL   GEOMETRY. 


365.  Corollary  to  364.  Parallel  sects  included 
between  parallel  planes  are  equal. 

Ex.  432.  If  two  ||  planes  meet  two  ||  planes,  the  four 
meets  are  ||. 

Ex.  433.  If  a  ||  to  the  meet  aft,  then  a  \\  a  and  a  \\  ft. 

Ex.  434.   If  a  ||  ft,  aLa  is  ±ft. 

Ex.  435.  Through  A  draw  a  \\  ft.     [Solution  unique.] 

Ex.  436.  If,  in  a,  a  cross  a' ,  in  ft,  b  cross  b' ,  and  a  ||  b, 
a'  ||  b',  then  a  ||  ft. 

Ex.  437.  Through  A  all  st's  ||  a  are  coplanar. 

Ex.  438.  Two  planes  ||  to  a  third  are  ||. 

Ex.  439.  The  intercepts  on  ||s  between  a  and  a  \\  a  are  -», 

Ex.  440.   If  AB  |1  a  and  BC  ||  «,  then  plane  A BC  \\  a. 

Ex.  441.  If  three  sects  are  =  and  ||,  the  as  of  their 
adjoining  ends  are    =  and  ||. 

Ex.  442.  If  A  in  a  ||  a,  AB  \\  a  is  in  a. 

366.  Theorem.  //  two  angles  have  their  sides 
respectively  parallel  and  on  the  same  side  of  the 
straight  through  their  vertices y  they  are  equal. 


Fig.  140. 


Hypothesis.  AB\\A'Bf  with  A  and  A'  on  the 
same  side  of  BBf ;  also  CB  \\  C'B'  with  C  and  C 
on  same  side  of  BB'. 


GEOMETRY  OF  PLANES.  155 

Proof.  From  A  take  (by  66)  A  A1 1|  BB' ';  .\  (by 
95)  AA'  =  5£'  and  A '5'  =A£.  In  same  way  CC'H 
and  -  BB'  and  B'C  =  BC.  But  then  A  A'  -  CC  and 
(by  353)  AA'\\CC'\   :.  (by  100)  AC-A'C;    /.  (by 

58)  aabcsaa'S'c. 

367.  Corollary  to  366.  Parallels  intersecting  the 
same  plane  are  equally  inclined  to  it. 

368.  Definition.  Let  two  planes,  a,  /?,  intersect 
in  the  straight  a.  Let  A  and  A'  be  points  on  a. 
Erect  now  at  A  and  A'  perpendiculars  to  a  in  one 
hemiplane  a'  of  a,  and  also  in  hemiplane  /?'  of  p. 
Then  (by  366)  the  angle  of  the  perpendiculars  at 
A  is  equal  to  the  angle  of  the  perpendiculars  at  A'. 


Fig.   141. 

We  call  this  angle  the  inclination  of  the  two  hemi- 
planes  a'  and  /?'. 

When  the  inclination  is  a  right  angle  the  planes 
are  said  to  be  perpendicular  to  each  other. 

369.  Theorem.  //  a  straight  is  perpendicular  to 
a  given  plane,  any  plane  containing  this  straight 
is  perpendicular  to  the  given  plane. 

Proof.  At  the  foot  of  the  given  perpendicular 
erect  in  the  given  plane  a  perpendicular  to   the 


156  RATIONAL    GEOMETRY. 

meet  of  the  planes.  From  the  definition  of  a 
perpendicular  to  a  plane  (334)  the  given  perpen- 
dicular makes  with  this  a  r't  a£  ;  but  this  angle 
is  (by  368)  the  inclination  of  the  planes. 

Ex.  443.  A  plane  _L  the  meet  of  two  planes  is  JL  to 
each;    and  inversely. 

Ex.  444.  Through  a  in  a  draw  fi±a. 

369  (b).  Corollary  to  369.  A  straight  and  its  pro- 
jection on  a  determine  a  plane  perpendicular  to  a. 

370.  Theorem.  //  two  planes  are  perpendicular 
to  each  other,  any  straight  in  one,  perpendicular  to 
their  meet,  is  perpendicular  to  the  other. 

371.  Corollary  to  370.  If  two  planes  are  per- 
pendicular to  each  other,  a  straight  from  any  point 
in  their  meet,  perpendicular  to  either,  lies  in  the 
other. 

For  the  perpendicular  to  their  meet  in  one  is 
perpendicular  to  the  other,  and  (by  340)  there  is 
only  one  perpendicular  to  a  plane  at  a  point. 

Ex.  445.   If  A  in  a_L/?,  from  A,  a±fi  is  in  a. 
Ex.  446.   If  a  st'  be  ||  to  a  plane,  a  plane  _L   to  the  st' 
is  1  to  the  plane. 

372.  Corollary  to  371.  If  each  of  two  inter- 
secting planes  is  perpendicular  to  a  given  plane, 
their  meet  is  perpendicular  to  that  plane. 

Proof.  The  perpendicular  to  this  third  plane 
from  the  foot  of  the  meet  of  the  others  is  (by  371) 
in  both  of  them. 

Ex.  447.  Through  a  st*  ||  a  to  pass  /?  ||  a. 

Ex.  448.  Through  a  draw  «JL/9. 

Ex.  449.  Through  A  draw  a±p  and  j. 


GEOMETRY  OF  PLANES. 


57 


373.  Theorem.      //  two    straights   be  cut  by  three 
parallel  planes  the  corresponding  sects  are  proportional. 


Fig.  142. 

Let  A B,  CD  be  cut  by  the  parallel  planes  a, 
/?,  r  in  A,  E,  B  and  C,  F,  D. 

To  prove  AE:EB  =CF:FD. 

Proof.  If  AD  cut  0  in  Gt  then  (by  364)  EG  \\  BD 
and  AC\\GF. 

:.  (by  235)  AE:EB=AG:GD  and  ^(7:6^= 
CF:FL>. 

.'.  AE:EB=CF:FD. 

Ex.  450.  Investigate  the  inverse  of  373. 

374.  Theorem.  Two  straights  not  co planar  have 
one,  and  only  one,  common  perpendicular. 

Given  a  and  b  not  coplanar. 

To  prove  there  is  one,  and  only  one,  straight 
perpendicular  to  both. 

Proof.  Through  any  point  A  of  a  take  c\\b. 
Then  (by  361)  the  plane  ac  or  a  ||  b.     The  projection 


158 


RATIONAL    GEOMETRY. 


b'  of  b  on  a  cuts  a,  say  in  B' ;    else  were  a  ||  bf  \\  b. 
Then,  in  plane  b%  B'B  drawn  ±b'  is  (by  370)  ±a 


Fig.  143- 
and  (by  74)  also  ±b,  and  is  the  only  common  per- 
pendicular to  a  and  b.  For  any  common  per- 
pendicular meeting  a  at  B"  is  J_  b"  through  B"  ||  6, 
which  is  in  a,  and  .\  ±«,  hence  -B''  is  a  point  of 
6'  the  projection  of  b  on  a;  .'.  identical  with  Bf,  the 
cross  of  bf  with  a. 

375.  Corollary  to  374.  Their  common  perpen- 
dicular is  the  smallest  sect  between  two  straights 
not  coplanar. 

For  (by  142)  BB' <BA. 

Ex.  451.  No  st's  joining  points  in  two  non-coplanar 
st's  can  be  ||. 

Ex.  452.  From  A,  B,  C,  costraight,  are  dropped  to  a 
non-coplanar    st'     JLs    AD,     BE,     CF.     Prove    AB:BC 
=DE:EF. 

Ex.  453.  A  plane  _L  to  the  common  _L  to  two  st's  at 
its  bisection-point,  bisects  every  sect  from  one  st'  to  the 
other. 

Ex.  454.  Principle  of  Duality.  When  any  figure  is  given 
we  may  construct  a  dual  figure  by  taking  planes  instead 
of  points,  and  points  instead  of  planes,  but  straights  where 
we  had  straights. 

The  figure  dual  to  four  non-coplanar  points  is  four 
non-copunctal  planes.     State  the  dual  of  the  following: 

Two  planes  determine  a  straight. 


GEOMETRY  OF  PLANES.  159 

Three  non-costraight  planes  determine  a  point. 

A  straight  and  a  plane  not  on  it  determine  a  point. 

Two  straights  through  a  point  determine  a  plane. 

Ex.  455.  A  polygon  whose  vertices  are  always  necessarily 
coplanar  is  what? 

Ex.  456.  From  each  point  in  a  plane  costraight  with  and 
equally  inclined  to  two  others  the  two  J_s  to  them  are  =. 

Ex.  457.  If  two  planes  are  respectively  _L  to  two  others 
and  the  intersection  of  the  first  pair  ||  to  that  of  the  second 
pair,  the  inclinations  are    =  or  supplemental. 

Ex.  458.  Parallels  have  as  projections  on  "any  plane 
parallels  or  points. 

Ex.  459.  Parallel  sects  are  proportional  to  their  pro- 
jections on  a  plane. 

Ex.  460.  From  a  point,  is  to  two  planes  make  an 
^  =  or  supplemental  to  the  inclination  of  the  planes. 

Ex.  461.  A  st'  has  the  same  inclination  to  ||  planes. 

Ex.  462.  If  three  meets  of  three  planes  are  ||,  the  sum 
of  the  three  inclinations  is  two  r't  2£s. 

Ex.  463.  If  a  st'  is  ||  to  each  of  two  planes,  any  plane 
on  it  cuts  them  in  ||s. 

Ex.  464.  Through  a  point  without  two  non-coplanar  st's 
passes,  in  general,  a  sin  le  st'  cutting  both. 

Ex.  465.  Why  does  each  foot  of  a  three-legged  stool 
meet  the  floor  while  one  foot  of  a  four-legged  chair  may 
be  above  the  floor? 

Ex.  466.  If  a,  b  non-coplanar,  a  j_  a  meets  /?  ±  b  in  c  _L 
y  ||  a  and  b. 

Ex.  467.  If  two  projections  of  a  trio  of  points  on  two 
intersecting  planes  give  costraight  trios,  the  original 
three  are  costraight.     State  the  exception. 

Ex.  468.  No  oblique  to  a  plane  makes  equal  angles  with 
three  straights  in  the  plane. 

Ex.  469.  Draw  a  plane  with  the  same  inclination  to 
two  given  planes. 

Ex.  470.  Draw  a  straight  that  shall  cross  three  straights, 
no  two  coplanar. 


160  RATIONAL    GEOMETRY. 

Ex.  471.  Draw  a  st'  to  cross  two  given  non-coplanar 
st's  and  ||  to  another  given  st'.  [Show  solution,  in  general, 
unique.] 

Ex.  472.  In  a  plane  find  a  point  which  joined  with  three 
given  points  without  the  plane  gives  equal  sects. 

Ex.  473.  The  projection  on  a  of  r't  ^  (a,  6)  is  ar't^  if 
a  ||  a. 

Ex.  474.  If  a  perpendicular  to  a  plane  be  projected 
on  any  second  plane  this  projection  is  at  right  angles  to 
the  meet  or  intersection  of  the  planes. 

Ex.  475.  From  a  point  without  a  plane,  if  there  be 
drawn  the  perpendicular  to  the  plane  and  also  a  perpen- 
dicular to  a  straight  in  the  plane,  the  join  of  the  feet  of 
these  perpendiculars  is  at  right  angles  to  the  straight. 

Ex.  476.  Two  planes  being  given  perpendicular  to  each 
other,  draw  a  third  perpendicular  to  both. 

Ex.  477.  Three  planes,  no  two  parallel,  either  intersect 
in  one  point  (are  copunctal)  or  in  one  straight  (are  co- 
straight)  or  have  their  three  intersection-straights  (meets) 
parallel. 

Ex.  478.  If  two  straights  be  at  right  angles  either  is 
in  the  plane  through  their  point  of  intersection  (cross) 
perpendicular  to  the  other. 

Ex.  479.  If  three  planes  have  two  of  their  intersec- 
tion-straights parallel,  the  third  is  parallel  to  both. 

Ex.  480.  All  straights  on  two  intersecting  straights, 
but  not  on  their  cross,  are  coplanar. 

Ex.  481.  If  the  vertices  of  a  triangle  give  equal  sects 
when  joined  to  a  point  without  their  plane,  the  foot  of 
the  perpendicular  from  this  point  to  the  plane  is  the  tri- 
angle's circumcenter. 

Ex.  482.  All  points  which  joined  to  three  given  points 
give  three  equal  sects  are  where? 

Ex.  483.  All  coplanar  points  which  joined  to  a  given 
point  give  equal  sects  are  where? 

Ex.  484.  If  a  plane  contains  one  straight  perpendicular 
to  a  second  plane,  every  straight  in  the  first  plane  per- 


GEOMETRY  OF  PLANES.  161 

pendicular  to  the   intersection-straight  (the  meet)  of  the 
planes  is  also  perpendicular  to  the  second. 

Ex.  485.  Any  plane  is  equally  inclined  to  two  parallel 
planes. 

Ex.  486.  If  jls  from  a  point  to  two  intersecting  planes 
be  =  it  determines  with  their  meet  a  plane  equally  in- 
clined to  them. 

Ex.  487.  Construct  a  plane  containing  a  given  straight 
and  perpendicular  to  a  given  plane. 

Ex.  48S.  Two  perpendiculars  from  a  point  to  two  inter- 
secting planes  determine  a  plane  perpendicular  to  the 
meet  of  the  two  planes. 

Ex.  489.  If  each  of  three  planes  be  perpendicular  to 
the  other  two,  their  three  meets  are  also  perpendicular 
to  the  planes  and  to  one  another. 

Ex.  490.  If  any  number  of  planes  perpendicular  to  a 
given  plane  have  a  common  point,  they  have  a  common 
meet  (intersection-straight). 

Ex.  491.  If  the  meets  of  several  planes  are  parallel,  the 
perpendiculars  to  them  from  any  given  point  are  coplanar. 

Ex.  492.  Perpendiculars  from  two  vertices  of  a  parallelo- 
gram to  a  plane  through  the  other  two  are  equal. 

Ex.  493.  Two  sides  of  an  equilateral  triangle  are  equally 
inclined  to  any  plane  through  the  third. 

Ex.  494.  If  two  straights  be  not  coplanar,  find  a  point 
in  one  which,  joined  to  two  given  points  in  the  other, 
gives  equal  sects. 

Ex.  495.   If  «  _1_  p  and  r.  and  the  meet  «/3  ||  ay,  then  p  \\y. 

Ex.  496.  If  the  four  sides  of  a  quadrilateral  be  not  co- 
planar it  is  called  skew.  No  three  sides  of  a  skew  quad- 
rilateral are  coplanar,  nor  can  its  four  ^s  be  r't. 

Ex.  497.  If  a  sect  divide  one  pair  of  opposite  sides  of  a 
skew  quadrilateral  proportionally,  and  another  divide 
the  other  pair  in  another  proportion,  these  two  sects  will 
cross  and  each  cut  the  other  as  it  cuts  the  sides. 

Ex.  498.  Find  that  point  in  a  given  plane  from  which  the 
sum  of  the  sects  to  two  given  points  on  the  same  side  is  least. 


1 62  RATIONAL   GEOMETRY. 

Ex.  499.  If  A  and  a  are  in  a,  and  B  not,  find  the  point 
in  a  from  which  the  sum  of  sects  to  A  and  B  is  least. 

Ex.  500.  Every  plane  not  on  a  vertex  cuts  an  even 
number  of  the  sides  of  a  skew  quad'  internally  and  an 
even  number  externally. 

Ex.  501.  If  non-coplanar  as  ABC,  A'B'C  have  A  A', 
BB' ,  CO  copunctal,  then  the  three  pairs  of  sides  AB, 
A'B'\  AC,  A'C'\  BC,  B'C  intersect  in  three  costraight 
points. 

Ex.  502.  If  two  st's  in  one  plane  be  equally  inclined 
to  another  plane  they  make  =  ^s  with  the  common  sec- 
tion of  these  planes. 

Ex.  503.  If  three  planes  be  each  _L  to  the  other  two, 
the  sq'  of  the  sect  from  their  common  intersection  to  an- 
other point  equals  the  sum  of  the  sq's  of  the  three  ±s 
from  that  point  to  the  planes. 

Ex.  504.  If  three  st's  be  each  J,  to  the  other  two,  twice 
the  sq'  of  the  sect  from  their  common  intersection  to 
another  point  equals  the  sum  of  the  sq's  of  the  three  _l_s 
from  that  point  to  the  st's. 

Ex.  505.  Draw  a  st'  to  cut  three  given  non-intersecting 
st's  so  that  the  intercepts  may  be  as  two  given  sects. 

Ex.  506.  If  a  plane  cut  a  tetrahedron  in  a  ||g'm,  the 
plane  is  ||  to  two  opposite  edges. 

Ex.  507.  The  aggregate  of  all  points  is  divided  by  four 
planes  into  (in  general)  fifteen  regions. 

Ex.  508.  The  medians  of  a  skew  quadrilateral  bisect  one 
another. 

Ex.  509.  If  two  medians  of  a  skew  quadrilateral  be  _|_  the 
diagonals  are  = ,  and  sections  ||  to  them  are  llg'ms  of  =  pe- 
rimeter. 


CHAPTER  XII. 

POLYHEDRONS  AND  VOLUMES. 

Polyhedrons. 

376.  Definition.  A  tetrahedron  is  the  figure  con- 
stituted by  four  non-coplanar  points,  their  sects 
and  triangles. 

The  four  points  are  called  its  summits,  the  six 
sects  its  edges,  the  four  triangles  its  faces.  Every 
summit  is  said  to  be  opposite  to  the  face  made  by 
the  other  three;  every  edge  opposite  to  that  made 
by  the  two  remaining  summits. 

A  point  is  within  the  tetrahedron  if  it  is  within 
any  sect  made  by  any  summit  and  a  point  within 
its  opposite  face.  Points  not  within  or  on  are 
without. 

The  faces  taken  together  are  called  the  surface 
of  the  tetrahedron. 

377.  A  polyhedron  is  the  figure  formed  by  n  plane 
polygons  such  that  each  side  is  common  to  two. 

The  polygons  are  called  its  faces,  and  taken 
together,  its  surface.  Their  sects  are  its  edges; 
their  vertices  its  summits. 

A  convex  polyhedron  is  one  through  no  edge  of 
which  pass  more  than  two  faces,  and  which  has  no 
summits  on  different  sides  of  the  plane  of  a  face. 

163 


1 64  RATIONAL   GEOMETRY. 

A  polyhedron  of  five,  six,  eight,  twelve,  twenty 
faces  is  called  a  pentahedron,  hexahedron,  octa- 
hedron, dodecahedron,  icosahedron. 

378.  A  pyramid  is  a  polyhedron  of  which  all  the 
faces,  except  one,  are  copunctal.  This  one  face 
is  called  the  base,  and  the  summit  not  on  it  the 
apex. 

The  faces  which  meet  at  the  apex  are  called 
lateral  faces,  and  together  the  lateral  surface;  the 
edges  meeting  at  the  apex  are  called  lateral  edges. 

The  perpendicular  from  the  apex  to  the  plane 
of  the  base  is  called  the  altitude  of  the  pyramid. 

379.  Euler's  Theorem.  In  any  convex  polyhedron 
the  number  of  faces  increased  by  the  number  of  sum- 
mits exceeds  by  two  the  number  of  edges. 

To  prove  F  +  S  =  E  +  2. 

Proof.  Let  e  be  any  edge  joining  the  summits 
A,  B  arid  the  faces  a,  /?;  and  let  e  vanish  by  the 
approach  of  B  to  Ai  If  a  and  /?  are  neither  of 
them  triangles,  they  both  remain,  though  reduced 
in  rank  and  no  longer  collateral,  and  the  poly- 
hedron has  lost  one  edge  e  and  one  summit  B. 

If  p  is  a  triangle  and  a  no  triangle,  fi  vanishes 
with  e  into  an  edge  through  A,  but  a  remains. 
The  polyhedron  has  lost  two  edges  of  /?,  one  face  /?, 
and  one  summit  B. 

If  /?  and  a  are  both  triangles,  /?  and  a  both  vanish 
with  ey  five  edges  forming  those  triangles  are  re- 
duced to  two  through  A,  and  the  polyhedron  has 
lost  three  edges,  two  faces,  and  the  summit  B. 

In  any  one  of  these  cases,  whether  one  edge  and 
one  summit  vanish,  or  two  edges  disappear  with 


POLYHEDRONS   AND    VOLUMES.  165 

a  face  and  a  summit,  or  three  edges  with  a  summit 
and  two  faces,  the  truth  or  falsehood  of  the  equa- 
tion 

F+S=E+2 

remains  unaltered. 

By  causing  all  the  edges  which  do  not  meet  any 
face  to  vanish,  we  reduce  the  polyhedron  to  a 
pyramid  upon  that  face.  Now  the  relation  is  true 
of  the  pyramid ;  therefore  it  is  true  of  the  undimin- 
ished polyhedron. 

380.  Theorem.  The  sum  of  the  face  angles  of 
any  convex  polyhedron  is  equal  to  four  right  angles 
taken  as  many  times,  less  two,  as  the  polyhedron  has 
summits. 

To  prove  I  =  (S-2)4  r't  if. 

Proof.  Since  E  denotes  the  number  of  edges, 
2E  is  the  number  of  sides  of  the  faces. 

Taking  an  exterior  angle  at  each  vertex,  the 
sum  of  the  interior  and  exterior  angles  is  2E2  r't  ^ , 
or  £4  r't  ^ .  But  the  exterior  angles  of  each  face 
make  4  r't  $  ;  ,\  the  exterior  angles  of  F  faces 
make  F4  r't  ^  . 

.-.  J-£4rt«*  -F4r't  4  =(£-F)4r't  *. 

But  (by  379)  -F  +  S  =  E+2; 

.\E-F  =  S-2) 

.'.  I  =  (S-2)4v't  4. 

Ex.  510.  The  number  of  face  angles  in  the  surface  of 
any  polyhedron  is  twice  the  number  of  its  edges. 

Ex.  t;n     Tf  a  polyhedron   has  for  faces  only  polygons 


1 66  RATIONAL   GEOMETRY. 

with  an  odd  number  of  sides,  it  must  have  an  even  num- 
ber of  faces. 

Ex.  512.  If  the  faces  of  a  polyhedron  are  partly  of  an 
even,  partly  of  an  odd,  number  of  sides,  there  must  be  an 
even  number  of  odd-sided  faces. 

Ex.  513.  The  number  of  face  angles  on  a  polyhedron 
can  never  be  less  than  thrice  the  number  of  faces. 

Ex.  514.   In  every  polyhedron  §S<_E. 

Ex.  515.   In  any  polyhedron  E+6<^S. 

Ex.  516.   In  any  polyhedron  E  +6<$F. 

Ex.  517.   In  every  polyhedron  E<$S. 

Ex.  518.   In  every  polyhedron  E<$F. 

Ex.  519.  In  a  polyhedron,  not  all  the  summits  are 
more  than  five  sided;  nor  have  all  the  faces  more  than 
five  sides. 

Ex.  520.  There  is  no  seven-edged  polyhedron. 

Ex.  521.  For  every  convex  polyhedron  the  sum  of  the  face 
angles  is  four  times  as  many  right  angles  as  the  difference 
between  the  number  of  edges  and  faces. 

Ex.  522.  How  many  regular  convex  polyhedrons  are 
possible  ? 

Ex.  523.  In  no  polyhedron  can  triangles  and  three- 
faced  summits  both  be  absent;  together  are  present  at 
least  eight. 

Ex.  524.  A  polyhedron  without  triangular  and  quad- 
rangular faces  has  at  least  twelve  pentagons;  a  poly- 
hedron without  three-faced  and  four-faced  summits  has 
at  least  twelve  five-faced. 

The  volume  of  tetrahedrons  and  polyhedrons. 

381.  Theorem.  The  product  of  an  altitude  of  a 
tetrahedron  by  the  area  of  its  base  is  independent  of 
what  summit  one  chooses  as  apex. 

Proof.  From  H  and  H',  feet  of  altitudes  from 
D  and  C,  drop  perpendiculars  HKy  and  H'K'  to 
AB. 


POLYHEDRONS  AND   VOLUMES.  167 

Then  KDLAB.    [If  two  planes  ABC  and  HKD 
are  at  r't  ^  's,   then  a  st'  AK  in  one  _L  to  their 


Fig.  144. 
intersection  KH  is  also  J_  to  the  other  HDK  and 

In  same  way  AK'lK'C. 
.(by  366)  IK^IK'. 

.  the  r't  triangles  //1CX)  and  H'K'C  are  similar. 
.DK:CK'=DH:CH'. 
DK-CH'^CK'-DH. 
±AB •  DAT 'CH'  -=\AB. CK'  ■  DH. 

382.  Definition.  One-third  the  product  of  base 
and  altitude  of  a  tetrahedron  T  is  called  the  volume 
of  tetrahedron  T  and  designated  by  V(T). 

383.  Convention.  A  plane  through  an  edge  of 
a  tetrahedron  and  a  point  of  the  opposite  edge 
is  called  a  transversal  plane;  this  cuts  the  tetra- 
hedron into  two  tetrahedra  with  common  altitude 
whose  bases  are  coplanar.  Such  a  partition  is 
called  a  transversal  partition  of  the  tetrahedron. 


1 68  RATIONAL    GEOMETRY. 

384.  Theorem.  The  volume  of  any  tetrahedron 
is  equal  to  the  sum  of  the  volumes  of  all  the  tetrahedra 
which  arise  from  the  first  by  making  successively  a 
set  of  transversal  partitions. 

Proof.  From  the  distributive  law  in  our  sect- 
calculus  follows  immediately  that  the  volume  of 
any  tetrahedron  is  equal  to  the  sum  of  the  volumes 
of  two  tetrahedra  which  come  from  the  first  by 
any  transversal  partition.  Thus  if  the  given  tetra- 
c  hedron  A  BCD  is  cut  by 

/\\  the  transversal  plane  ACE 

/     I  \  \  passing  through  the  edge 

/         \\\  AC,    the   two    tetrahedra 

/  I    \     \        so    obtained,   AEBC  and 

/  I     \       \      AEDC,  have  in  common 

^0~~~~~^J       \  /       the    altitude  he  from    C. 
^v         I    ~y£  Moreover,  the  area  of  the 

^\IS  triangle  ABD  is  equal  to 

B  the  sum  of  the  areas  of 

FlG-  I45-  AEB  and  AED. 

Thus   V(TJ  +  V(T2)  =  JV^(A«)  +  iK-M*2)  = 

Now  our  theorem  follows  merely  by  repeated 
application  of  this  single  result. 

385.  Theorem.  However  a  tetrahedron  is  cut  by 
a  plane,  this  partition  can  be  obtained  in  a  set  of 
transversal  partitions  using  not  more  than  two  other 
planes* 

Proof.     Passing   the    case   in   which    the    plane 

*See  G.  Veronese,  Atti  del  R.  Istituto  Veneto,  t.  vi,  s.  vii; 
1894-95- 


POLYHEDRONS  AND    VOLUMES. 


169 


itself   goes    through    an    edge    of    the  tetrahedron 
A  BCD,  there  remain  three  cases: 

I.  The  plane  passes  through  a  single  summit, 
for  example,  A,  thus  cutting  the  tetrahedron  A  BCD 
into  the  tetrahedron  ABXY  and  the  pyramid 
with  quadrilateral  base  AXYCD. 


But  this  partition  is  in  the  set  of  transversal 
partitions  obtained  by  taking  successively  the 
planes  ADY  and  AYX. 

II.  The  plane  cuts  the  three  edges  copunctal 
in  a  summit,  for  example,  A,  thus  cutting  the 
tetrahedron  A  BCD  into  the  tetrahedron  AXYZ 
and  the  convex  polyhedron  XYZBCD.  But  this 
partition  is  in  the  set  of  transversal  partitions 
obtained  by  taking  successively  the  planes  BDY, 
BYZ,  YZX. 

III.  The  plane  cuts  two  pairs  of  opposite  edges, 
for  example,  A B,  CD,  and  AC,  BD. 

Thus  the  tetrahedron  is  cut  into  the  two  poly- 


170 


RATIONAL    GEOMETRY. 


hedrons  ADWXYZ  and  BCWXYZ.  Draw  the 
two  planes  BDX  and  ACZ.  By  these  at  the  same 
time  the   polyhedron   ADWXYZ  is   cut   into  the 


Fig.   148. 

three  tetrahedra  XZAYy  XZAD,  XZDWy  and 
the  polyhedron  BCWXYZ  cut  into  the  three 
tetrahedra  XZBY,  XZBC,  XZCW 


POLYHEDRONS  AND    VOLUMES.  I71 

But  that  these  six  tetrahedra  form  a  set  ob- 
tainable by  transversal  partitions  is  seen  by  taking 
first  the  plane  BDX,  and  then  in  the  tetrahedron 
BDXC  successively  the  planes  CXZ  and  ZXW '; 
in  the  tetrahedron  BDXA  successively  the  planes 
AXZ  and  YZX. 

386.  Theorem.  //  a  tetrahedron  T  is  in  any  way 
cut  into  a  certain  finite  number  of  tetrahedra  7\., 
then  is  always  the  volume  of  the  tetrahedron  T 
equal  to  the  sum  of  the  volumes  of  all  the  tetra- 
hedra Tk. 

Proof.  The  plane  a  of  any  one  face  of  any 
one  of  the  tetrahedra  Tkl  say,  Tv  makes  in  T  a 
partition  which,  by  385,  can  be  obtained  in  a  set 
of  transversal  partitions  of  T  made  by  introducing 
two  other  planes  /?,  r,  cutting  T  into  Tt.  When 
a  cuts  any  tetrahedron  Tk,  say  T2,  add  in  this  T2 
the  two  planes  d,  c,  making  the  corresponding  set 
of  transversal  partitions  in  this  T2. 

When  /?  cuts  one  of  these  new  tetrahedra, 
say  T2a,  add  in  it  the  requisite  two  planes  £,  77. 
So  do  for  any  tetrahedron  T3  met  by  /?.  Then 
in  the  same  way  successively  for  y. 

Now  produce  a  second  face  of  Tv  say  d,  to  cut 
those  tetrahedra  Tt  in  which  this  face  is  situated. 
Add  in  each  of  these  Tt  the  two  planes  to  make 
the  transversal  partition.  When  any  plane  cuts 
a  tetrahedron  already  existing,  add  the  requisite 
two  planes. 

Now  produce  a  third  face  of  T1  to  the  nearest 
tetrahedron  trans versally  obtained  from  T.  Finally 
take  in  the  fourth  face  of   1\.     Then   7\  appears 


172 


RATION A L    GEOMETRY. 


in  a  set  of  transversal  partitions  of  T;  while  if 
cut  itself,  it  is  by  a  set  of  transversal  partitions 
of  Tv     In  the  same  way  for  every  other  Tk. 

Thus  after  a  finite  number  of  constructions  we 
reach  by  a  set  of  transversal  partitions  of  la  final 
set  of  tetrahedra  Tf,  which  are  at  the  same  time 
also  reached  by  transversal  partitions  of  the  tetra- 
hedra Tk. 

3866.  Of  the  fundamental  theorem  386: 

For  any  partition  of  a  tetrahedron  into  tetrahedra 
the  sum  of  their  volumes  equals  its  volume,  the  follow- 
ing alternative  proof  is  due  to  S.  0.  Schatunovsky 
of  Odessa. 

386c.  Partition  method  I.  Cut  a  face,  for  ex- 
ample   BCD    (Fig.    149),    of    the    tetrahedron    in 


Fig.   149. 

question  A  BCD  into  a  finite  number  n  of  triangles, 
and  join  their  vertices  with  the  summit  A.  The 
tetrahedra  so  obtained  have  a  common  summit  A, 


POLYHEDRONS  AND   VOLUMES. 


173 


and  the  faces  opposite  it  coplanar.  [For  the  sake 
of  distinctness,  Fig.  149  shows  only  one  such  part- 
tetrahedron.] 

By  280,  the  area  of  the  base  of  the  original  tetra- 
hedron equals  the  sum  of  the  areas  of  all  the  bases 
of  the  part-tetrahedra.  It  and  they  have  the  same 
altitude.  Our  multiplication  of  sects  is  distributive. 
Therefore  its  volume  equals  the  sum  of  theirs. 

386J.  This  I  contains  transversal  partition  ($&$) 
as  a  special  case  [i.e.  for  11  =  2]. 

3866?.  Partition  method  II.  Cut  the  tetrahedron 
A  BCD  so  that  all  summits  of  the  part-tetrahedra 
lie  on  three  edges  meeting  in  the  same  apex,  for 
example  on  the  edges  AB,  AC,  AD  (Fig.  150). 


150. 


Then  on  the  base  BCD  are,  besides  the  points  B, 
C,  D}  no  summits  of  the  part-tetrahedra;  and  con- 
sequently the  face  BCD  is  a  face  of  a  part-tetra- 
hedron. The  fourth  summit  E  of  this  part-tetra- 
hedron may  lie  on  the  edge  AB  (Fig.  150). 


174  RATIONAL   GEOMETRY. 

Therefore  this  partition  into  n  part-tetrahedra 
may  be  obtained  by  first  cutting  the  tetrahedron 
A  BCD  by  a  transversal  partition  (here  through 
DC)  into  two  tetrahedra  BCDE  and  ECDA,  and 
then  ECDA  in  the  same  way  into  n  —  i  part-tetra- 
hedra. 

386/.  Partition  method  III.  Cut  the  given  tetra- 
hedron into  part-tetrahedra  by  I,  and  these  each 
into  part-tetrahedra  by  II. 

386^.  Partition  method  IV.  {Partition  with  help 
of  Central  Projection.) 

Take  a  point  0  either  without  the  tetrahedron 
ABCD  or  coincident  with  one  of  its  summits,  for 
example  A,  and  let  the  rays  OA,  OB,  OC,  OD 
meet  a  plane  a  in  A' ,  B' ,  C,  D'  which  cannot  be 
costraight  since  A,  B,  C,  D  are  not  coplanar.  Call- 
ing A'  the  central  projection  of  A,  then  the  figure 
made  by  the  central  projections  of  the  edges  of  the 
tetrahedron  ABCD  is  in  general  (the  special  cases 
are  hereafter  exhaustively  treated)  a  quadrilateral 
with  its  two  diagonals,  which  cut  it  into  triangles. 
Cut  now  these  last  triangles  in  any  way  into  part- 
triangles,  altogether  n  in  number.  The  vertices 
of  these  part-triangles  (in  our  figures  always  only 
one  such  triangle  is  shown)  join  with  the  projection- 
centre  0,  whereby  n  tetrahedra  OT/,  0T2',  .  .  .  QTJ 
are  made,  which  have  a  common  summit  0,  and 
whose  bases  T/,  T2',  .  .  .  Tn'  are  all  the  part-tri- 
angles of  the  projection-figure  A'B'C'D' .  If  0 
coincides  with  A  then  the  edges  of  the  tetrahedra 
0T*/,  0T2'y  .  .  .  0Tnf  cut  the  original  tetrahedron 
ABCD  into  n  part-tetrahedra. 


POLYHEDRONS  AND   VOLUMES.  175 

On  the  other  hand,  if  0  is  different  from  A,  then 
the  edges  of  07/,  07/,  .  0Tn'  cut  the  tetra 
hedron  A  BCD  into  a  number  of  truncated  tetra- 
hedra, among  which  as  special  cases  may  appear 
the  pyramid  of  five  summits  and  the  complete 
tetrahedron.  These  pyramids  appear  if  a  vertex 
of  the  corresponding  triangle  T1  falls  on  one  of 
the  sects  A'B\  A'C  .  .  .  Complete  tetrahedra 
appear  if  two  vertices  of  the  triangle  7'  fall  on  one 
of  the  said  sects. 

Always  cut  these  truncated  tetrahedra  into  three 
(these  pyramids  into  two)  complete  tetrahedra, 
using  a  diagonal  of  each  quadrilateral  face  [as  in 

385  HI. 

By  this  last  partition  we  get  no  new  summits. 

The  partition  of  the  original  tetrahedron  into 
part-tetrahedra  so  attained  is  called  Partition  with 
the  aid  of  central  projection.  The  different  cases 
may  be  more  explicitly  set  forth  as  follows: 

Case  1.  0  coincides  with  A.  This  is  Partition 
method  I  [see  386c]. 

Case  2.  0  lies  on  the  prolongation  of  an  edge,  for 
example  DA   (Fig.   151). 

The  projection-points  A'  and  D'  coincide  and  we 
get  on  the  plane  a  as  projection  of  A  BCD  a  tri- 
angle A'B'C. 

That  face  of  the  truncated  tetrahedron  (or  pyra- 
mid of  five  summits  or  complete  tetrahedron)  ob- 
tained by  the  above  given  construction,  which  by 
the  projection  of  Tmr  is  made  in  the  face  BCD 
designate  by  Tm\   that  in  face  ABC  by  TIm. 

The    tetrahedra    obtained    by    the    transversal 


176  RATIONAL   GEOMETRY. 

partition  of  this  truncated  tetrahedron  designate 
respectively  by  tj,  tj't  tj".  Then  is  the  volume  of 
A  BCD  equal  to  the  sum  of7  the  volumes  of  all  the  t. 
For  every  tetrahed  on  0Tm  is  divided  into  at 
most  four  part-tetrahedra  0TIm,  tj}  tm",  tj",  all 


Fig.  151. 
of  whose  summits  lie  on  the  three  edges  meeting 
in  apex  0,  that  is  every  0Tm  is  divided  by  Parti- 
tion method  II.     Therefore 

V(OTt)  -  V(OTn)  +  V(*/)  +  V(t»)  +  V(V") 
V(OT2)  -  V{OTl2)  +  V(W)+  V&")  +  V(tn, 


V(OTn)  m  V{OTln)  +  V(U)  +  V(tn")  +  V(tn>")- 
Adding  these  equations  to  one  another  and  notic- 
ing that  (by  3866)  on  the  one  side 

V{OTx)+  V(OT2)  +  . .  .  +  V{OTn)  =  V(OBCD), 
on  the  other  side 

V(OTh)+V(OTl2)  +  .  . .  +  V(OTIn)  =  V(OABC), 


POLYHEDRONS  AND   VOLUMES. 


177 


we  find,  using  I  to  mean  the  sum  from  w  =  i  up 

to  n=n, 

V(OBCD)  =  ViOABC)  +  T[V(tn')  +  V(tn")  +  ^(C")J- 

n<=  1 

Furthermore,  the  division  of  the  tetrahedron 
OBCD  by  the  plane  ABC  into  the  tetrahedra  OABC 
and  ABCD  is  a  transversal  partition,  and  so  (by  384) 

V(OBCD)  -  V(OA£C)  +  F(yl5CL>). 

The  last  two  equations  give  finally 

V (ABCD)  =  "T\y(tJ)  +  V(tn")  +  V(tH'")] ;    . 

n  =  i 

therefore  also  for  this  case  our  theorem  is  proven. 

Case  3 .  0  lies  without  the  tetrahedron  on  one  of  its 
boundary  planes,  that  is  in  the  plane  of  one  of  its 
faces;  for  example,  on  the  plane  ABD  (Fig.  152). 


Fig.  152. 
Then  the  projections  A',  B' ,  D'  are  costraight; 
therefore  one  of  them,   say  A'f  lies   between    the 
other  two. 


178  RATIONAL    GEOMETRY 

B' ,  C,  D'  make  a  triangle  which  is  cut  by  the 
sect  A'C  into  two  part-triangles  A'B'C  and  A'C'D'. 
Correspondingly  the  plane  OA'C  cuts  the  tetra- 
hedron ABCD  into  two  part-tetrahedra  AAt  BC, 
and  AAfiD\  and  since  this  plane  goes  through  the 
edge  AC,  therefore  it  makes  a  transversal  partition, 
and  we  have  (by  384) 

V(ABCD)  =  V(AAXBC)  +  V(AAfD). 

Since,  however,  0  lies  on  the  prolongation  of  the 
edge  ALA,  therefore  (by  Case  2) 

V{AAlBC)=IV{tb) 
and 

V{AA,CD)=lV{td), 

where  tb  are  all  the  tetrahedra  into  which  in  accord- 
ance with  the  above  method  the  tetrahedron 
AAXBC  is  divided  [and  td  those  in  AALCD]. 

These  last  three  equations  give  now  V(ABCD)  = 
IV(t),  where  t  are  all  the  part-tetrahedra  of 
ABCD. 

Case  4.  If  0  lies  on  no  one  of  the  boundary  planes 
of  the  tetrahedron  ABCD,  then  no  three  of  the  pro- 
jection-points A',  B',  C,  D'  are  ccstraight. 

Consider  first  the  case  in  which  one  of  these  points 
(say  Af)  falls  within  the  triangle  B'C'D'  made  by 
the  other  three  (Fig.  153). 

The  joining  sects  A'B',  A'C',  A'D'  cut  the  tri- 
angle B'C'D'  into  three  part-triangles,  and  corre- 
spondingly the  tetrahedron  ABCD  is  divided  into 
three    part-tetrahedra    AAJBC.    AAJ3D,    AA.CD, 


POLYHEDRONS  AND   VOLUMES. 


179 


giving   a   case   of    Partition   method   I ;    therefore 

(by  386c) 

V(ABCD)  -  V(AAXBC)  +  V(AAXBD)  +  V(AAXCD). 

If  we  now,  as  above,  divide  the  triangles  A'B'C\ 
A'B'D' ,   A' CD'  into   part-triangles,   project  these 


Fig.  153. 

back  upon  the  corresponding  part-tetrahedra,  and 
designate  the  part-tetrahedra  obtained  in  the 
above  given  way  of  the  tetrahedra  AAXBC,  AAXBD, 
AAXCD  with  td,  tc,  tb  respectively,  then  we  obtain 
the  three  equations 

V{AAXBC)  -IV(t4) ;  V{AAXBD)  =  IV(tc) ; 
V(AAxCD)=ZV(tb), 

since  0  lies  on  the  ray  AXA,  that  is  we  have  here 
each  time  Case  2. 

These  last  four  equations  give  now  V(ABCD)  ■■= 
IV (t),  where  /  are  all  the  part-tetrahedra  of  ABCD. 


iSo 


RATIONAL   GEOMETRY. 


Case  5.  If,  finally,  again  0  lies  in  no  one  of  the 
boundary  planes  of  the  tetrahedron  A  BCD,  but  each 
of  the  projections  A',  Br ,  C,  D'  falls  without  the 
triangle  made  by  the  remaining  projection-points, 
then  the  projections  of  the  edges  of  the  tetrahedron 
make  a  convex  quadrilateral  with  its  two  diagonals. 
This  quadrilateral  is  divided  by  its  diagonals  into 
four  triangles,  M'A'C,  M'A'D'%  M'B'C\  M'B'D' 
(Fig.  154). 


Fig.  154. 

The  plane  OA'B'  divides  the  tetrahedron  by  a 
transversal  partition  into  two  parts,  so  that  we 
have  (by  384) 

V(ABCD)  -  V(AMBC)  +  V(AMBD)9 

where  M  is  the  point  corresponding  to  the  inter- 
section point  M'  of  the  diagonals  A'Br  and  CD'. 


POLYHEDRONS  AND    VOLUMES.  181 

If  we  now  divide  each  triangle  M'A'C,  M'A'D', 
M'B'C,  M'B'D'  into  part-triangles  and  project 
back  upon  the  faces  of  the  tetrahedron  A  BCD, 
then  the  tetrahedra  AMBC  and  AMBD  are  divided 
each  into  part-tetrahedra,  which,  in  general,  may 
be  designated  respectively  tc  and  td.  Since  now, 
moreover,  0  lies  in  the  plane  AMB,  therefore 
(by  Case  3)  we  have  the  equations 

V{AMBC)=lV(tc)    and    V(AMBD)  =  IV(td). 
Consequently  is  also  in  this  case 
V(ABCD)  =  IV(t). 

This  completes  the  proof  of  the  theorem  that  in 
every  partition  of  a  tetrahedron  by  central  projec- 
tion the  sum  of  the  volumes  of  the  part-tetrahedra 
equals  the  volume  of  the  whole  tetrahedron. 

386/1.  The  most  general  partition  can  be  built  up 
from  the  four  partition  methods  already  given,  and 
this  proves  the  fundamental  theorem  386. 

For  let  A  BCD  be  a  tetrahedron  and  Plt  P2,  .  .  . 
Pk  the  part-tetrahedra  which  arise  from  any  par- 
tition of  it. 

If,  now,  we  project  all  these  tetrahedra  from  the 
point  A  upon  the  face  BCD,  then  their  projections, 
which  necessarily  all  fall  within  the  triangle  BCD, 
overlie  and  overlap,  in  general,  manifoldly,  and 
cut  one  another  into  polygons.  When  we  cut  these 
polygons  into  triangles  and  join  their  vertices 
with  A  we  divide  each  tetrahedron,  Pm(m  =  i, 
2,  .  .  .  k),  into  a  number  of  truncated  tetrahedra 
(including  perhaps  pyramids  of  five  summits  and 


1 82  RATIONAL    GEOMETRY. 

complete  tetrahedra)  which  in  turn  in  the  well- 
known  way  we  divide  into  further  part-tetrahedra. 
Since  the  so  obtained  partition  of  each  part- 
tetrahedron  Pm  into  further  part-tetrahedra,  which 
may  be  designated  tm't  tm"y  .  .  .  *£«>,  is  accom- 
plished with  the  aid  of  central  projection,  for  each 
projection-center  lies  without  each  tetrahedron, 
only  that  one  with  the  summit  A  excepted,  so  is 

V{Pm)  -  V{tm')  +  F(V')  + .  .  .  +  V(^J), 

If  we  now  give  m  successively  the  values  i, 
2,  .  .  .  k,  we  obtain  k  such  equations,  which  added 
give  the  following: 

But,  on  the  other  hand,  every  tetrahedron  AT, 
where  T  is  a  part-triangle  of  BCD,  cuts  out  from 
the  aggregate  of  part-tetrahedra  t  a,  set,  and  each 
tetrahedron  of  this  set  appears  once  and  only  once 
in  the  above  sum. 

At  the  same  time  all  summits  of  these  last  part- 
tetrahedra  lie  on  the  three  edges  from  A  of  the 
particular  tetrahedron  AT;  that  is,  AT  is  divided 
by  this  set  of  tetrahedra  according  to  Partition 
method  II. 

Furthermore,  the  whole  tetrahedron  ABCD  is 
divided  into  tetrahedra  AT  according  to  Partition 
method  I,  so  that  it  is  divided  according  to  Parti- 
tion method  III  into  part-tetrahedra  t. 

Now   this   complex   of  tetrahedra   /  is  identical 


POLYHEDRONS  AND   VOLUMES  183 

with  the  complex  t$t  where  /  =  i,  2,  .  .  .  nm\  m  =  i, 
2,  .  .  .  k. 
Consequently 

V(ABCD)-iV(Wiz;j::  *.)• 

which  combined  with  the  previous  equation  gives 
the  desired  proof  of  the  fundamental  theorem 

m  =  k 

V(ABCD)  =  2  V(Pm). 

m=i 

387.  Theorem.  If  a  polyhedron  P  is  cut  into 
tetrahedra  in  two  different  ways,  then  the  sum  of  the 
volumes  of  the  tetrahedra  of  the  first  partition  equals 
that  of  the  second. 

Proof.  Suppose  P  divided  into  m  tetrahedra 
hs  t2,  .  .  .  tm,  and  again  into  n  tetrahedra  */,  t2\ 
.  .  .  U. 

Construct  a  tetrahedron  T  which  contains  the 
polyhedron  P,  and  cut  the  polyhedron  bounded 
by  the  surface  of  P,  and  that  of  T  in  any  definite 
way  into  tetrahedra  3T/,  TV,  .  .  . 

Thus  we  obtain  two  partitions  of  the  tetrahe- 
dron T  and  (by  386)  the  equations 

V(T)=V(tl)  +  V(Q  +  ... 

+  V(tm)  +  V(Tl')  +  V(T,')  +  ... 

V(T)=V(tl')  +  V(t2')  +  ... 

+  V(tn')  +  V(Tl')  +  V(T2')  +  ... 
whence 

V(tt)  +  V(t2)  +  ...  +  V(tm) 

=  V(t1')  +  V(t2')  +  ...  +  V(tJ). 


l84  RATIONAL    GLOME  TRY 

388.  Definition.  The  volume  of  a  polyhedron  is 
the  sum  of  the  volumes  of  any  set  of  tetrahedra 
into  which  it  is  cut. 

389.  Definition.  Two  polyhedra  P  and  Q  of 
equal  volume  are  said  to  be  equal.  P  is  called 
greater  than  Q  if  the  volume  of  P  is  greater  than 
the  volume  of  Q ;  less  if  volume  is  less. 

UV(P)=V(Q),  we  say  P^Q. 
UV(P)>V(Q),  we  say  P>Q. 
IiV(P)<V(Q),  we  say  P<Q. 
The  three  cases  are  mutually  exclusive. 

390.  Corollary.  If  a  polyhedron  be  cut  into  poly- 
hedra, the  sum  of  their  volumes  equals  its  volume. 

391.  Corollary.  If  a  polyhedron  be  cut  into 
polyhedra,  if  one  of  these  be  omitted  it  is  not 
possible  with  the  others,  however  arranged,  to 
make  up  the  original  polyhedron. 

The  Prismatoid  Formula. 

392.  Definition.  A  prismatoid  is  a  polyhedron 
having  for  base  and  top  any  two  polygons  in  parallel 
planes,  and  whose  lateral  faces  are  triangles  deter- 
mined by  the  vertices  so  that  each  lateral  edge 
with  the  succeeding  forms  a  triangle  with  one  side 
of  the  base  or  of  the  top. 

The  altitude  of  a  prismatoid  is  the  perpendicular 
from  top  to  base. 

A  number  of  different  prismatoids  thus  have 
the  same  base,  top,  and  altitude. 

A  prismatoid  with  a  point  as  top  is  a  pyramid. 
If  both  base  and  top  of  a  prismatoid  are  sects,  it 
is  a  tetrahedron. 


POLYHEDRONS  AND   VOLUMES.  185 

If  a  side  of  the  base  and  a  side  of  the  top  which 
form  with  the  same  lateral  edge  two  sides  of  two 
adjoining  faces  are  parallel,  then  these  two  tri- 
angular faces  fall  in  the  same  plane,  and  together 
form  a  trapezoid. 

393.  A  prismoid  is  a  prismatoid  whose  base  and 
top  have  the  same  number  of  sides,  and  every 
corresponding  pair  parallel. 

394.  A  frustum  of  a  pyramid  is  a  prismoid  with 
base  and  top  similar. 

395.  Corollary.  Every  prismoid  with  triangular 
base  is  the  frustum  of  a  pyramid. 

396.  A  section  of  a  prismatoid  is  the  polygon 
determined  by  a  plane  perpendicular  to  the  alti- 
tude. 

397.  Theorem.  The  area  of  a  section  of  a  pyramid 
is  to  the  area  of  the  base  as  the  square  of  the  perpendic- 
ular on  it,  from  the  apex,  is  to  the  square  of  the  alti- 
tude of  the  pyramid. 

To  prove  S:B=p2:a2. 

Proof.  The  section  and  base 
are  similar,  since  corresponding 
diagonals  cut  them  into  tri- 
angles similar  in  pairs  because 
having  all  their  sides  respec- 
tively proportional,  each  corre- 
sponding  pair  being  as  a  lateral 
edge  to  the  sect  on  it  made  by 
apex  and  section,  which  in  turn  FlG-  x55- 

are  as  altitude  to  perpendicular  on  section.     But 
(by  300)  the  areas  are  as  the  squares  of  these. 

398.  To  find  the  volume  of  any  pyramid. 


1 86  RATIONAL    GEOMETRY. 

Rule.     Multiply  one-third  its  altitude  by  the  area 

of  its  base.     Formula,  Y=-B. 

3 

Proof.  Cutting  this  base  into  triangles  by  planes 
through  the  apex,  we  have  a  partition  of  the  given 
pyramid  into  triangular  pyramids  (tetrahedra)  of 
the  same  altitude  whose  bases  together  make  the 
polygonal  base. 

399.  To  find  the  volume  of  any  prismatoid. 

Rule.  Multiply  one -fourth  its.  altitude  by  the 
sum  of  the  base  and  three  times  a  section  at  two- 
thirds  the  altitude  from  the  base. 

Formula,  D  =-  (B  +  35). 
4 


Proof.  Any  prismatoid  may  be  divided  into 
tetrahedra,  all  of  the  same  altitude  as  the  pris- 
matoid; some,  as  CFGO,  having  their  apex  in  the 
top  of  the  prismatoid  and  their  base  within  its 
base;  some,  as  OABC,  having  three  summits 
within  the  top  and  the  fourth  in  the  base  of  the 


POLYHEDRONS  AND   VOLUMES. 


187 


prismatoid,  thus  having  for  base  a  point  and  for 
top  a  triangle;  and  the  others,  as  ACOG,  having 
for  base  and  top  a  pair  of  opposite  edges,  a  sect 
in  the  plane  of  the  base  and  a  sect  in  the  plane  of 
the  top  of  the  prismatoid,  as  OG  and  AC. 


Fig.   157. 

Therefore  if  the  formula  holds  good  for  tetra- 
hedra  in  these  three  positions,  it  holds  for  the 
prismatoid,  their  sum. 

In  (1)  call  5  the  section  at  two-thirds  the  alti- 
tude from  the  base  B{ ;  then  5t  is  %a  from  the  apex. 
Therefore  (by  397)  the  areas 


.-.  A  =^  +  3^)  -fa  +  iBJ  =$aBlf 


which  (by  382)  equals  Ylt  the  volume  of  this  tetra- 
hedron. 

In  (2)  the  base  B2  is  a  point,  and  S2  is  \a  from 


1 88  RATIONAL   GEOMETRY. 

this    point,    which    is    the    apex    of    an     inverted 
pyramid. 

•'•  (by 397)  the  areas52:T2  =  (fa)2:a2;  .\S2=|T2> 

'•  D2=-(B2  +  3S2)  =-(o  +  iT2)  =iaT2  =  Y2. 
4  4 

In  (3)  let  KLMN  be  the  section  S3.  Now  the 
areas 

AANK:AAGO=AN2-^G2^(^a)2:a2  =  i:g; 
AGNM:AGAC  =  GN2:GA2  =  (%a)2:a2==  4:9. 

But  the  whole  tetrahedron  D3  and  the  pyramid 
CANK  may  be  considered  as  having  their  bases 
in  the  same  plane,  AGO,  and  the  same  altitude, 
a  perpendicular  from  C: 

.\CANK:D3=  a  ANK:  a  AGO  =  1:9; 
:.CANK=\D3. 

In  the  same  way 

OGNM :D3  =  aGNM :  aGAC  =  4:9; 

.• .  CA iVX  +  O^A^M  =  {P, ; 
.• .  CKLMN  +  OKLMN  -  *D9. 

But  by  (398)  OKLMN  +  OKLMN  =  i-iaSt  + 
i  •  § aS3  =  Ja53 ;     / .  ^L>3  =  JaS3 ; 

/.£3  =  ^3^(£3  +  3S3), 
4  4 

since  here  the  area  B3  =  o. 


POLYHEDRONS  AND    VOLUMES.  189 

400.  Corollary  to  399.  Since,  in  a  frustum  of  a 
pyramid,  B  and  5  are  similar;  .*.  if  b  and  s  be  corre- 
sponding edges, 

2\ 

B:S  =  b2:s2;     /.the  volume  F=-b(i+^A. 

401.  Definition.  A  prism  is  a  prismatoid  whose 
base  and  top  are  congruent. 

A  right  prism  is  one  whose  lateral  edges  are 
perpendicular  to  its  base. 

A  par allelo piped  is  a  prism  whose  bases  are 
prallelograms. 

A  cuboid  is  a  parallelopiped  whose  six  faces 
are  rectangles. 

A  cube  is  a  cuboid  whose  six  faces  are  squares. 

402.  Corollary  to  400.  To  find  the  volume  of 
any  prism. 

Rule.  Multiply  its  altitude  by  the  area  of  its  base. 
Formula,   V(P)=a-B. 

.-.  To  find  the  volume  of  a  cuboid. 

Rule.  Multiply  together  any  three  copunctal  edges, 
that  is,  its  length,  breadth,  and  thickness. 

j  03.  A  cube  whose  edge  is  the  unit  sect  has  for 
volume  this  unit  sect,  since  1X1X1=1. 

Any  polyhedron  has  for  volume  as  many  such 
unit  sects  as  the  polyhedron  contains  such  cubes 
on  the  unit  sect. 

The  number  expressing  the  volume  of  a  poly- 
hedron will  thus  be  the  same  in  terms  of  our  unit 
sect,  or  in  ,terms  of  a  cube  on  this  sect,  considered 
as  a  new  kind  of  unit,  a  unit  solid.  Such  units, 
though  traditional,  are  unnecessary. 


190  RATIONAL    GEOMETRY, 

Ex.  525.  If  the  altitude  of  the  highest  Egyptian  pyramid 
is  138  meters,  and  a  side  of  its  square  base  228  meters, 
find  its  volume. 

Ex.  526.  The  pyramid  of  Memphis  has  an  altitude  of 
73  Toises;  the  base  is  a  square  whose  side  is  116  Toises. 
If  a  Toise  is  1-95  meters,  find  the  volume  of  this  pyramid. 

Ex.  527.  A  pyramid  of  volume  15  has  an  altitude  of 
9  units.     Find  the  area  of  its  base. 

Ex.  528.  Find  the  volume  of  a  rectangular  prismoid 
of   12   meters  altitude,   whose  top  is   5   meters  long  and 

2  meters  broad,   and  base   7   meters  long  and  4  meters 
broad. 

Ex.  529.  In  a  prismoid  15  meters  tall,  whose  base  is 
36  square  meters,  each  basal  edge  is  to  the  top  edge  as 

3  to  2.     Find  the  volume. 

Ex.  530.  Every  regular  octahedron  is  a  prismatoid  whose 
bases  and  lateral  faces  are  all  congruent  equilateral  tri- 
angles.    Find  its  volume  in  terms  of  an  edge  b. 

Ex.  531.  The  bases  of  a  prismatoid  are  congruent 
squares  of  side  b,  whose  sides  are  not  parallel;  the  lateral 
faces  are  eight  isosceles  triangles.     Find  the  volume. 

Ans.  £a62(2  +  2*). 

Ex.  532.  If  from  a  regular  icosahedron  we  take  off 
two  five-sided  pyramids  whose  vertices  are  opposite 
summits,  there  remains  a  solid  bounded  by  two  congruent 
regular  pentagons  and  ten  equilateral  triangles.  Find 
its  volume  from  an  edge  b.  Ans.  %b\$+  2(5)*]. 

Ex.  533.  Both  bases  of  a  prismatoid  of  altitude  a  are 
squares;  the  lateral  faces  isosceles  triangles.  The  sides 
of  the  upper  base  are  parallel  to  the  diagonals  of  the 
lower  base,  and  half  as  long  as  these  diagonals;  and  b  is 
a  side  of  the  lower  base.     Find  the  volume.     Ans.  £a&2. 

Ex.  534.  The  upper  base  of  a  prismatoid  of  altitude 
a  =  6  is  a  square  of  side,  62  =  7-07107;  the  lower  base  is  a 
square  of  side  6,  =  10,  with  its  diagonals  parallel  to  sides 
of  the  upper  base;  the  lateral  faces  are  isosceles  triangles. 
Find  volume. 

Ex.  535.  Every  prismatoid  is  equal  in  volume  to  three 


POLYHEDRONS  AMD  VOLUMES.  191 

pyramids  of  the  same  altitude  with  it,  of  which  one  has 
for  base  half  the  sum  of  the  prismatoid's  bases,  and  each 
of  the  others  its  mid-cross  section: 

D  =  *a(~T^+2M)  =MB1+4M+B2). 

Ex.  536.  If  a  prismatoid  have  bases  with  angles  re- 
spectively equal  and  their  sides  parallel,  in  volume  it 
equals  a  prism  plus  a  pyramid,  both  of  the  same  altitude 
with  it,  whose  bases  have  the  same  angles  as  the  bases  of 
the  prismatoid,  but  the  basal  edges  of  the  prism  are  half 
the  sum,  and  of  the  pyramid  half  the  difference,  of  the 
corresponding  sides  of  both  the  prismatoid's  bases. 

Ex.  537.  If  the  bases  of  a  prismatoid  are  trapezoids 
whose  mid-sects  are  bx  and  b2,   and  whose  altitudes  are 

bl  +b2     1  ax  —a2  bx  —b2\ 


at  and  a2,  the  volume  =  a(    ■ — ■ 


Ex.  538.  A  side  of  the  base  of  a  frustum  of  a  square 
pyramid  is  25  meters,  a  side  of  the  top  is  9  meters,  and 
the  height  is  240  meters.  Required  the  volume  of  the 
frustum. 

Ex.  539.  The  sides  of  the  square  bases  of  a  frustum 
are  50  and  40  centimeters.  Each  lateral  edge  is  30  centi- 
meters.    How  many  liters  would  it  contain? 

Ex.  540.  In  the  frustum  of  a  pyramid  whose  base  is 
50  and  altitude  6,  the  basal  edge  is  to  the  corresponding 
top  edge  as  5  to  3.     Find  volume. 

Ex.  541.  Near  Memphis  stands  a  frustum  whose  height 
is  142-85  meters,  and  bases  are  squares  on  edges  of  185-5 
and  3-714  meters.     Find  its  volume. 

Ex.  542.  In  the  frustum  of  a  regular  tetrahedron, 
given  a  basal  edge,  a  top  edge,  and  the  volume.  Find 
the  altitude. 

Ex.  543.  A  wedge  of  10  centimeters  altitude,  4  centi- 
meters edge,  has  a  square  base  of  36  centimeters  perimeter. 
Find  volume. 

Ex.  544.  The  diagonal  of  a  cube  is  n.       Find  its  volume. 

Ex.  545.  The  edge  of  a  cube  is  n.  Approximate  to 
the  edge  of  a  cube  twice  as  large. 


19 2  RATIONAL   GEOMETRY. 

Ex.  546.  Find  the  cube  whose  volume  equals  its  super- 
ficial area. 

Ex.  547.  Find  the  edge  of  a  cube  equal  to  three  whose 
edges  are  a,  b,  I. 

Ex.  548.  If  a  cubical  block  of  marble,  of  which  the 
edge  is  1  meter,  costs  one  dollar,  what  costs  a  cubical  block 
whose  edge  is  equal  to  the  diagonal  of  the  first  block? 

Ex.  549.  If  the  altitude,  breadth,  and  length  of  a  cuboid 
be  a,  b,  I,  and  its  volume  V, 

(1)  Given  a,  b,  and  superficial  area;   find  V. 

(2)  Given  a,  b,  V;    find  /. 

(3)  Given  V,  (ab),  (bl);  find  I  and  b. 


(4)   Given  V,   l-j-J  ,  l-j  J  ;   find  a  and  b. 


(5)   Given  (ab),  (al),  (bl);   find  a  and  b. 

Ex.  550.  If  97  centimeters  is  the  diagonal  of  a  cuboid 
with  square  base  of  43  centimeters  side,  find  its  volume. 

Ex.  551.  The  volume  of  a  cuboid  whose  basal  edges 
are  12  and  4  meters  is  equal  to  the  superficial  area.  Find 
its  altitude. 

Ex.  552.  In  a  cuboid  of  360  superficial  area,  the  base 
is  a  square  of  edge  6.     Find  the  volume. 

Ex.  553.  A  cuboid  of  volume  864  has  a  square  base 
equal  in  area  to  the  area  of  two  adjacent  sides.  Find 
its  three  dimensions. 

Ex.  554.  In  a  cuboid  of  altitude  8  and  superficial  area 
160  the  base  is  square.     Find  the  volume. 

Ex.  555-  The  volume  of  a  cuboid  is  144,  its  diagonal 
13,  the  diagonal  of  its  base  5.     Find  its  three  dimensions. 

Ex.  556.  In  a  cuboid  of  surface  108,  the  base,  a  square, 
equals  in  area  the  area  of  the  four  sides.     Find  volume. 

Ex.  557.  What  is  the  area  of  the  sheet  of  metal  re- 
quired to  construct  a  rectangular  tank  (open  at  top)  12 
meters  long,  10  meters  broad,  and  8  meters  deep? 

Ex.  558.  The  base  of  a  prism  to  meters  tall  is  an 
isosceles  right  triangle  of  6  meters  hypothenuse.  Find 
volume. 


POLYHEDRONS  AND   VOLUMES.  19  3 

Ex.  559.  In  a  prism  the  area  of  whose  base  is  210  the 
three  sides  are  rectangles  of  area  336,  300,  204.  Find 
volume. 

Ex.  560.  A  right  prism  whose  volume  is  480  stands 
upon  an  isosceles  triangle  whose  base  is  10  and  side  13. 
Find  altitude. 

Ex.  561.  In  a  right  prism  whose  volume  is  54,  the  lat- 
eral area  is  four  times  the  area  of  the  base,  an  equilateral 
triangle.     Find  basal  edge. 

Ex.  562.  The  vertical  ends  of  a  hollow  trough  are 
parallel  equilateral  triangles  with  1  meter  in  each  side, 
a  pair  of  sides  being  horizontal.  If  the  length  between 
the  triangular  ends  be  6  meters,  find  the  volume  of  water 
the  trough  will  contain. 


CHAPTER  XIII. 

TRIDIMENSIONAL  SPHERICS. 

404.  Definition.  If  C  is  any  given  point,  then 
the  aggregate  of  all  points  A  for  which  the  sects 
CA  are  congruent  to  one  another  is  called  a  sphere. 
C  is  called  the  center  of  the  sphere,  and  CA  the 
radius. 

Every  point  B,  such  that  CA  >  CB  is  said  to  be 
within  the  sphere.  If  CA  <  CD,  then  D  is  without 
the  sphere. 

405.  Theorem.  Any  ray  from  the  center  of  a 
sphere  cuts  the  sphere  in  one,  and  only  one,  point. 

406.  Theorem.  Any  straight  through  its  center 
cuts  the  sphere  in  two,  and  only  two,  points. 

407.  Definition.  A  sect  whose  end-points  are 
on  the  sphere  is  called  a  chord. 

408.  Definition.  Any  chord  through  the  center 
is  called  a  diameter.  Its  end-points  are  called 
opposite  points  of  the  sphere. 

409.  Theorem.  Every  diameter  is  bisected  by 
the  center. 

410.  Corollary  to  106  and  409.  A  plane  through 
its  center  meets  the  sphere  in  a  circle  with  radius 
equal  to  that  of  the  sphere.  Such  a  circle  is  called 
a  great  circle  of  the  sphere. 

194 


TRIDIMENSIONAL  SPHERICS.  195 

4i  i.  Corollary  to  410.  All  great  circles  of  the 
sphere  are  congruent,  since  each  has  for  its  radius 
the  radius  of  the  sphere. 

412.  Theorem.  Any  two  great  circles  of  a  sphere 
bisect  each  other. 

Proof.  Since  the  planes  of  these  circles  both 
pass  through  the  center  of  the  sphere,  therefore 
on  their  intersection  is  a  diameter  of  the  sphere 
which  is  a  diameter  of  each  circle. 

413.  Theorem.  //  any  number  of  great  circles  pass 
through  a  given  point,  they  will  also  pass  through  the 
opposite  point. 

Proof.  The  given  point  and  the  center  of  the 
sphere  determine  the  same  diameter  for  each  of 
the  circles. 

414.  Corollary  to  413.  Through  opposite  points 
an  indefinite  number  of  great  circles  can  be  passed. 

415.  Theorem.  Through  any  two  non-opposite 
points  on  a  sphere,  one,  and  only  one,  great  circle 
can  be  passed. 

Proof.  For  the  two  given  points  and  the  center 
of  the  sphere  determine  its  plane. 

416.  Definition.  A  straight  or  plane  is  called 
tangent  to  a  sphere  when  it  has  one  point,  and 
only  one,  in  common  with  the  sphere. 

Two  spheres  are  called  tangent  to  each  other 
when  they  have  one  point,  and  only  one,  in  com- 
mon. 

417.  Theorem.  A  straight  or  plane  having  the 
foot  of  the  perpendicular  to  it  from  the  center  in  com- 
mon with  the  sphere  is  tangent. 

Proof.     This  perpendicular,  a  radius,  is  (by  142) 


196  RATIONAL   GEOMETRY. 

less  than  any  other  sect  from  the  center  to  this 
straight  or  plane.  Therefore  every  point  of  the 
straight  or  plane  is  without  the  sphere  except  the 
foot  of  this  radius. 

418.  Theorem.  If  a  straight  has  a  given  point  not 
the  foot  of  the  perpendicular  to  it  from  the  center 
in  common  with  a  sphere,  it  has  a  second  point  on 
the  sphere. 

Proof.  This  is  the  other  end-point  of  the  sect 
from  the  given  point  bisected  by  this  perpen- 
dicular. 

419.  Theorem.     //  a  plane  has  a   point   not    the 

©foot  of  the  perpendicular  to  it 
from  the  center  in  common  with  a 
sphere,  it  cuts  the  sphere  in  a 
circle. 
Proof.  If  A  be  the  common 
point  and  C  the  foot  of  the 
perpendicular,  the  circle  OC(CA) 
Fig.  158.  is  on  the  sphere. 

420.  Corollary  to  419.  The  straight  through 
the  center  of  any  circle  of  a  sphere  perpendicular 
to  its  plane  passes  through  the  center  of  the  sphere. 

421.  Definition.  The  two  opposite  points  in 
which  the  perpendicular  to  its  plane,  through  the 
center  of  a  circle  of  the  sphere,  meets  the  sphere, 
are  called  the  poles  of  that  circle,  and  the  diameter 
between  them  its  axis 

422.  Theorem.  Any  three  points  on  a  sphere  deter- 
mine a  circle  on  the  sphere  (I  3  and  419). 

423.  Theorem.  The  radius  of  any  circle  of  the 
sphere  whose   plane   does  not  contain   the   center 


TRIDIMENSIONAL   SPHERICS.  197 

of  the  sphere  is  less  than  the  radius  of  a  great 
circle. 

Proof.     The  hypothenuse  (by  142)  is  >   a  side. 

424.  Definition.  A  circle  on  the  sphere  whose 
plane  does  not  contain  the  center  of  the  sphere  is 
called  a  small  circle  of  the  sphere. 

425.  Inverse  of  417.  Every  straight  or  plane 
tangent  to  the  sphere  is  perpendicular  to  the 
radius  at  the  point  of  contact.  For  if  not  it  would 
have  (by  418)  another  point  on  it. 

426.  Theorem.  //  two  spheres  have  two  points 
in  common  they  cut  in  a  circle  whose  center  is  in 
their  center-straight  and  whose  plane  is  perpendicular 
to  that  straight. 

Hypothesis.  Let  C  and  0  be  the  centers  of 
the  spheres  having  the  points  A  and  B  in  common. 

Conclusion.  They  have  in  common  all  points, 
and  only  those,  on  a  circle  with  center  on  OC  and 
plane  _L  to  OC. 


Fig.  159. 

Proof.  Since,  by  58,  aACO=aBCO,  /.perpen- 
diculars from  A  and  B  upon  OC  are  equal  and 
meet  OC  at  the  same  point,  D.  Thus  all,  but  only, 
points  like  A  and  B,  in  a  plane  J_  to  OC,  and  points 
of  OD(DA),  are  on  both  spheres. 


198 


RATIONAL    GEOMETRY. 


427.  Corollary  to  426.  If  two  spheres  are  tan- 
gent, either  internally  or  externally,  their  centers 
and  point  of  contact  are  costraight. 

428.  Theorem.  Four  points,  not  coplanar,  deter- 
mine a  sphere. 

Proof.  Let  A,  B,  C,  D 
be  the  four  given  points. 
Then  (by  352)  the  plane 
a  J_  to  and  bisecting  A  B 
meets  /?  ±  to  and  bisecting 
BC  in  EH  1  ABC,  and 
meets  dl_  to  and  bisecting 
BD  in  FOLABD. 

.\EH±EGy  and  F01.FG, 
and  (by  77)  EH  and  FO 
meet,  say,  at  0;  .*.  (by 
346)  0  is  one,  and  the 
only  center  of  a  sphere  containing  A,  B,  C,  D. 

429.  Corollary  to  428.  The  four  perpendiculars 
to  the  faces  of  a  tetrahedron  through  their  circum- 
centers,  and  the  six  planes  bisecting  at  right  angles 
the  edges,  are  copunctal  in  its  circumcenter. 

430.  Problem.  To  inscribe  a  sphere  in  a  given 
tetrahedron. 

Construction.  Through  any  edge  and  any  point 
from  which  perpendiculars  to  its  two  faces  are 
equal,  take  a  plane.  Likewise  with  the  other 
edges  in  the  same  face.  The  cointersection  of 
these  three  planes  is  the  incenter  required. 

431.  Theorem.  The  sects  joining  its  pole  to  points 
on  any  circle  of  the  sphere  are  equal. 

Proof  (343). 


Fig.  160. 


TRIDIMENSIONAL   SPHERICS. 


r99 


432.  Corollary  to  431.  Since  equal  chords  have 
congruent  minor  arcs,  .'.  the  great-circle-arcs  join- 
ing a  pole  to  points  on  its 
circle  are  congruent.  Hence 
if  C  is  any  point  in  a  sphere 
a,  then  the  aggregate  of  all 
points  A  in  a,  for  which  the 
great-circle-arcs  CA  are  con- 
gruent to  one  another  is  a 
circle.  1  Fig.  161. 

433.  Theorem.     The   great-circle-arc   joining   any 
point  in  a  great  circle  with  its  pole  is  a  quadrant. 

Proof      The  angle  at  the  center  is  right. 

434.  Theorem.  If  A,  B  are 
non- opposite,  the  point  P  is  a 
pole  of  their  great  circle  when  the 
arcs  PA,  PB  are  both  great- 
circle -quadrants. 

For  each  of  the  angles  POA, 
POB  is  right  and  /.  PO±OAB. 

435.  Definition.  The  angle 
between  two  great-circle-arcs  on  a  sphere,  called 
a  spherical  angle,  is  the  angle  between  tangents 
to  those  arcs  at  their  point  of  meeting. 

436.  A  spherical  angle  is  the  inclination  of  the 
two  hemiplanes  containing  the  arcs. 

437.  Theorem.  Any  great  circle  through  a  pole 
0}  a  given  great  circle  is  perpendicular  to  the  given 
great  circle. 

Proof.     Their  planes  (by  369)  are  at  right  angles. 

438.  Inverse  of  437.     Any  great  circle  perpen- 


Fig.  162. 


200 


RATIONAL   GEOMETRY. 


dicular  to  a  given  great  circle  will  pass  through 
its  poles. 

439.  Theorem.  If  a  sphere  be  tangent  to  the  par- 
allel planes  containing  opposite  edges  of  a  tetrahe- 
dron, and  sections  made  in  the  sphere  and  tetra- 
hedron by  one  plane  parallel  to  these  are  of  equal 
area,  so  are  sections  made  by  any  parallel  plane. 


D 

Vr: c        "" 


!>-- 

r -&b 

1  \     /*/ 

s                       /         1 

\                ^V   /        / 

T\i    ^/f 

Fig.  163. 

Hypothesis.  Let  KJ  be  the  sect  _L  to  the  edges 
EF  and  GH  in  the  ||  tangent  planes.  Then  KJ  = 
DT,  the  diameter. 

Let  Ql(IP)  =  M0,  sections  made  by  the  plane 
±DT  at  I  and  IK  J  at  R,  where  KR  =DI. 

Let  ABCLSN  be  any  parallel  plane  through  a 
point  A  of  the  sphere. 

Conculsion.     LN  =  qC(CA). 
Proof.     Since 

a  LEU- a  MEW,     and     a  LHV~  A  MHZ, 
.:  MW:LU=EM  :EL  =JR  :JS  (by  373). 
MZ  :LV  =  HM:HL=KR:KS. 
But  (by  366)  $ZMW=ifVLU. 
;.  (by  299)  area  MO  :area  LN  =  WM-MZ  :UL-LV 

=  JR-RK:JS:SK. 


TRIDIMENSIONAL  SPHERICS.  201 

But  (by  325) 

area  el(IP)  :area  oC(CA)  =PI2:AC2 
-77 -ID :TC -CD  (by  245); 

.*.  area  LN  =  area  OC(CVI). 

440.  Cavalieri's  assumption.  If  the  two  sec- 
tions made  in  two  solids  between  two  parallel 
planes  by  any  parallel  plane  are  of  equal  area, 
then  the  solids  are  of  equal  volume. 

441.  Theorem.  The  volume  of  a  sphere  of  radius 
r  is  f  nr3. 

Proof.  A  tetrahedron  on  edge,  and  a  sphere 
with  this  tetrahedron's  altitude  for  diameter,  have 
(by  439)  all  their  corresponding  sections  of  equal 
area,  if  any  one  pair  are  of  equal  area. 

Hence  (by  440)  they  are  of  equal  volume. 

•  '•  (by  399)  vol.  sphere  =  JaS. 
.    But  a  =  2r,  and  (by  245)  5  =  fr-|r-7r. 

.'.  Vol.  sphere  =  f  •2r-§r-|r-7r=|^r3. 

442.  Definition.  The  area  of  a  sphere  is  the 
quotient  of  its  volume  by  one-third  its  radius. 

Area  of  sphere  =  4^r2. 

443.  Corollary  to  324.  The  area  of  a  sphere  is 
quadruple  the  area  of  its  great  circle. 

444.  Definition.  A  spherical  segment  is  the  piece 
of  a  sphere  between  two  parallel  planes.  If  one 
of  the  parallel  planes  is  tangent  to  the  sphere, 
the  segment  is  called  a  segment  of  one  base. 

445.  Corollary  to  439  and  440.     The  volume  of  a 

a 
spherical  segment  is  -7r(^i2  +  3^32),  where  r3  is  the 


202 


RATIONAL    GEOMETRY. 


radius  of  the  section  two-thirds  the  altitude  from 
the  base  whose  radius  is  rv  If  the  segment  is 
of  one  base  its  volume  is  \anr^\    which  in  terms 

of  r,  the  radius  of  the  sphere,  is  na2lr — ],  and 

equals  %na(r22-\ — ).  If  we  eliminate  r3  by  intro- 
ducing r2,  the  radius  of  the  top,  the  volume  of  the 
segment  is  inalsir^  +  r^)  +  a2]. 

446.  Problem.     Given    a    portion    of    a    sphere, 
find  its  radius. 

Construction.  Take  any  three  points  of  the  part 
given,  say  A,  B,  C.  The  plane  A,  B,  C  (by  419) 
cuts  the  sphere  in  a  circle.  The  straight  at  D, 
the  center  of  this  circle  perpendicular  to  the  plane 
ABC,  contains  the  center  0  of  the  sphere  (by  420) 
and  therefore  meets  the  sphere, 
say  at  P.  In  the  plane  PAD 
draw  AP'A_  to  AP  and  meet- 
ing DO  in  P'.  Bisect  PP'  in 
0.  Then  0  is  the  center  of 
the  sphere  and  OP  is  the  radius. 
Proof.  0  is  circumcenter  of 
PAPf. 

:.OP  =  OA.     But    since    OD 
is_LO(D)ZM  at  A  ;.OA=OB  =  OC. 

447.  Corollary  to  305. 


Fig.  164. 


OP 


DA2  +  DP: 
2DP 


that  is,  R  = 


r2+h'< 
2h 


Ex.  563.  A  circle  on  a  sphere  of  10  centimeters  radius 


TRIDIMENSIONAL   SPHERICS.  203 

has  its  center  8  centimeters  from  the  center  of  the  sphere. 
Find  its  radius. 

Ex.  564.  The  sects  from  the  centers  of  circles  of  equal 
area  on  a  sphere  to  the  center  of  the  sphere  are  equal. 

Ex.  565.  Where  are  the  centers  of  spheres  through 
three  given  points? 

Ex.  566.   Find  the  volume  of  a  sphere  whose  area  is  20. 

Ex.  567.  Find  the  radius  of  a  globe  equal  to  the  sum 
of  two  globes  whose  radii  are  3  and  6  centimeters. 

Ex.  568.  A  section  parallel  to  the  base  of  a  hemisphere, 
radius  1,  bisects  its  altitude.  Find  the  volume  of  each 
part. 

Ex.  569.  The  areas  of  the  parts  into  which  a  sphere  is 
cut  by  a  plane  are  as  5  to  7.  To  what  numbers  are  the 
volumes  of  these  parts  proportional? 

Ex.  570.  The  volume  of  a  spherical  segment  of  one 
base  and  height  8  is  1200.     Find  radius  of  the  sphere. 

Ex.  571.  Find  the  volume  of  a  segment  of  12  centi- 
meters altitude,  the  radius  of  whose  single  base  is  24 
centimeters. 

Ex.  572.  In  terms  of  sphere  radius,  find  the  altitude 
of  a  spherical  segment  n  times  its  base. 

Ex.  573.  Find  volume  of  a  spherical  segment  of  one 
base  whose  area  is  15  and  base  2  from  sphere  center. 

Ex.  574.  In  a  sphere  of  10  centimeters  radius  find  the 
radii  rt  and  r2  of  the  base  and  top  of  a  segment  whose 
altitude  is  6  centimeters  and  base  2  centimeters  from  the 
sphere  center. 

Ex.  575.  Out  of  a  sphere  of  12  centimeters  radius  is 
cut  a  segment  whose  volume  is  one-third  that  of  the 
sphere  and  whose  bases  are  congruent.  Find  the  radius 
of  the  bases. 

Ex.  576.  Find  the  radius  of  a  sphere  whose  area  equals 
the  length  of  a  great  circle. 

Ex.  577.  Find  the  volume  of  a  sphere  the  length  of 
whose  great  circle  is  n. 


204  RATIONAL   GEOMETRY. 

Ex.  578.  Find  the  radius  of  a  sphere  whose  volume 
equals  the  length  of  a  great  circle. 

Ex.  579.  The  volume  of  a  sphere  is  to  that  of  the  cir- 
cumscribing cube  as  n  to  6. 

Ex.  580.  Find  altitude  of  a  spherical  segment  of  one 
base  if  its  area  is  A  and  the  volume  of  the  sphere  V. 

Ex.  581.  The  radii  of  the  bases  of  a  spherical  segment 
are  5  and  4;  its  altitude  3.     Find  volume. 


CHAPTER  XIV. 

CONE  AND  CYLINDER. 

448.  Definition.  The  aggregate  of  straights  de- 
termined by  pairing  the  points  of  a  circle  each 
with  the  same  point  not  in  their  plane  is  called  a 
circular  cone  of  two  nappes. 

This  point  is  called  the  apex  of  the  cone.  Each 
straight  is  an  element. 

The  straight  determined  by  the  apex  and  the 
center  is  called  the  axis  of  the  cone. 

The  rays  of  the  cone  on  the  same  side  of  a  plane 
through  the  apex  perpendicular  to  the  axis  are 
one  nappe  of  the  cone. 

The  sects  from  the  apex  to  the  circle  are  often 
called  the  cone,  and  are  meant  when  we  speak  of 
the  area  or  the  volume  of  the  cone. 

When  each  element  makes  the  same  angle  with 
the  axis,  the  cone  is  called  a  right  cone. 

In  a  right  cone  all  sects  from  apex  to  circle  are 
equal,  and  each  is  called  the  slant  height. 

449.  Theorem.  Every  section  of  a  circular  cone 
by  a  plane  parallel  to  the  base  is  a  circle. 

Let  the  section  D'H'B'F'  of  the  circular  cone 
A-DHBF  be  parallel  to  the  base. 

To  prove  D'H'B'F'  a  circle. 

205 


2o6 


RATIONAL    GEOMETRY. 


Proof.  Let  C  be  the  center  of  the  base,  and 
C  the  point  of  the  axis  AC  in 
the  plane  D'H'B'.  The  plane 
through  AC  and  any  element  AB 
gives  radii  CB,  CD,  and  parallel 
to  them  the  sects  CB\  CD'. 

.'.(by  >]$)&ABC~&AB'C  and 
aACD-^aACD'. 
;.  (by    234)     CB'  :  CB=AC: 
Pig.  165,  AC-CD*  i  CD. 

But  CB=  CD.     .'.CB'=CD'. 
450-  Corollary  to  449. 

The  axis  of  a  circular  cone  passes  through  the 
center  of  every  section  parallel  to  the  base. 

451.  Theorem.  If  a  circular  cone  and  a  tetra- 
hedron have  equal  altitudes  and  bases  of  equal  area 
and  in  the  same  plane,  sections  by  a  plane  parallel 
to  the  bases  are  of  equal  area. 


Fig.  166. 
Proof.     BC  :  B'C-AC  :  AC  =AL  :  AD '.- 

-  VT  :  VT  -  VH  :  VH'  =GH  :  G'H'. 
.'.BC2  :B'C2  =  GH2  :  CH'2 
But  (by  325) 

area  oC(CB)  :  area  oC(CB') 


CONE  AND   CYLINDER.  207 

=  BC2  :  B'C'2=GH2  :  (TE'2 

=  area  &FGH  :  area  aF'G'H'  (by  300). 

But  by  hypothesis  area  OC(CB)  =area  aFGH. 
.-.  Area  oC'{C'B')  -area  aF'G'H' . 

452.  Corollary  to  451. 

Volume  of  circular  cone  is  (by  440)  =  volume  of 
tetrahedron  of  equal  altitude  and  base  =  -J ar.r2. 

453.  Theorem.  The  lateral  area  of  a  right  cir- 
cular cone  is  half  the  product  of  the  slant  height  by 
the  length  of  the  base. 

Proof.  It  has  the  same  area  as  a  sector  of  a 
circle  with  the  slant  height  as  radius  and  an  arc 
equal  in  length  to  the  length  of  the  cone's  base. 

•  ".  (by  323)  K  =  \ch  =  nrh. 

454.  Definition.  A  truncated  pyramid  or  cone  is 
the  portion  included  between  the  base  and  a  plane 
meeting  all  the  elements. 

A  frustum  of  a  cone  is  the  portion  included  be- 
tween the  base  and  a  plane  parallel  to  the  base. 

455.  Theorem.  The  lateral  area  of  a  frustum 
of  a  right  circular  cone  is  half  the  product  of  its  slant 
height  by  the  sum  of  the  lengths  of  its  bases. 

Proof.  It  is  the  difference  of  the  areas  of  two 
sectors  with  a  common 
angle,  the  lengths  of  the 
arcs  of  the  sectors  being 
equal  to  the  lengths  of 
bases  of  the  frustum. 

.*.  F  =  \h(cx  +  c2) 

=  7ih(rl  +  r2).  Fig.  167. 

456.  Corollary  to  451  and  399. 


2o8  RATIONAL    GEOMETRY. 

The  volume  of  the  frustum  of  a  circular  cone, 
T/.F  =  ia7r(r12  +  3r32), 

where  r3  is  the  radius  of  5. 

457.  Definition.  A  circular  cylinder  is  the  assem- 
blage of  straights  each  through  a  point  of  a  given 
circle  but  not  in  its  plane,  and  all  parallel. 

The  portion  of  this  assemblage  included  between 
two  planes  parallel  to  the  circle  is  also  called  a 
circular  cylinder.  The  sects  the  planes  cut  out 
are  called  the  elements  of  the  cylinder. 

The  two  circles  in  these  planes  are  called  the 
bases  of  the  cylinder. 

The  sect  joining  their  centers  is  called  the  axis. 

A  sect  perpendicular  to  the  two  planes  is  the 
altitude  of  the  cylinder. 

If  the  elements  are  perpendicular  to  the  planes, 
it  is  a  right  cylinder;   otherwise  an  oblique  cylinder. 

A  section  whose  plane  is  perpendicular  to  the  axis 
is  called  a  right  section  of  the  cylinder.  Any  two 
elements,  being  equal  and  parallel,  are  opposite 
sides  of  a  parallelogram;  hence  the  bases  and  all 
sections  parallel  to  them  are  equal  circles. 

A  truncated  cylinder  is  the  portion  between  a 
base  and  a  non-parallel  section. 

458.  Theorem.  The  volume  of  a  circular  cylin- 
der is  the  product  of  its  base  by  its  altitude. 

Proof.  If  a  prism  and  cylinder  have  equal 
altitudes  and  bases  of  equal  area,  any  sections 
parallel  to  the  bases  are  of  equal  area. 

.*.  (by  402)     V'C  =  a7ir2. 


CONE  AND   CYLINDER. 


209 


459.  The  lateral  area  of  a  circular  cylinder  is 
the  product  of  an  element  by  the  length  of  a  right 
section : 

C '  =  27tra. 


Fig.  168. 

Proof.  It  is  equal  to  the  area  of  a  parallelo- 
gram with  one  side  an  element  and  the  consecu- 
tive side  equal  to  the  length  of  a  base. 

An  altitude  of  this  parallelogram  equals  the 
length  of  the  right  section. 

460.  Corollary    to    459.     The    lateral    area    of   a 
truncated  circular  cylinder  is  the  prod- 
uct   of    the    intercepted    axis   by   the 
length  of  a  right  section. 

Proof.  For  substituting  an  oblique 
section  for  the  right  section  through 
the  same  point  of  the  axis  changes 
neither  the  area  nor  the  volume,  since 
the  portion  between  the  sections  is 
the  same  above  as  below  either. 

461.  Corollary  to  460.  The  volume  of  a  trun- 
cated circular  cylinder  is  the  product  of  the  inter- 
cepted axis  by  the  area  of  the  right  section. 

462.  Archimedes'  Theorem.  The  volume  of  a 
sphere  equals  two-thirds  the  volume  of  the  circum- 
scribed cylinder. 


Fig.  169. 


210  RATIONAL    GEOMETRY. 

Proof.  The  volume  of  the  circumscribed  cylin- 
der =  7rr2-2r  =  2  nr3. 

Ex.  582.  In  a  right  circular  cylinder  of  altitude  a, 
call  the  lateral  area  C  and  the  area  of  the  base  B. 

(1)  Given  a  and  C;   find  r. 

(2)  Given  B  and  C;    find  a. 

(3)  Given  C  and  a  =  2r;    find  C  +  25. 

(4)  Given  C+2B  and  a  =  r;   find  C. 

(5)  Given  a  and  B  +C,;   find  r. 

Ex.  583.  The  lateral  area  of  a  right  circular  cylinder 
is  equal  to  the  area  of  a  circle  whose  radius  is  a  mean 
proportional  between  the  altitude  of  the  cylinder  and  the 
diameter  of  its  base. 

Ex.  584.  In  area,  the  bases  of  a  right  circular  cylinder 
together  are  to  the  lateral  surface  as  radius   to  altitude. 

Ex.  585.  If  the  altitude  of  a  right  circular  cylinder 
is  equal  to  the  diameter  of  its  base,  the  lateral  area  is 
four  times  that  of  the  base. 

Ex.  586.  How  much  must  the  altitude  of  a  right  cir- 
cular cylinder  be  prolonged  to  increase  its  lateral  area  by 
the  area  of  a  base? 

Ex.  587.  The  lateral  area  of  a  right  circular  cone  is 
twice  the  area  of  the  base;   find  the  vertical  angle. 

Ex.  588.  Call  the  lateral  area  of  a  right  circular  cone 
K,  its  altitude  a,  the  basal  radius  r,  the  slant  height  h. 

(1)  Given  a  and  r;    find  K. 

(2)  Given  a  and  h;   find  K. 

(3)  Given  K  and  h;   find  r. 

Ex.  589.  How  much  canvas  is  required  to  make  a 
conical  tent  20  meters  in  diameter  and  12  meters  high? 

Ex.  590.  How  far  from  the  vertex  is  the  cross-section 
which  halves  the  lateral  area  of  a  right  circular  cone? 

Ex.  591.  Given  the  volume  and  lateral  area  of  a  right 
circular  cylinder;    find  radius. 

Ex.  592.  Given  lateral  area  and  altitude  of  a  right 
circular  cylinder;    find  volume. 


CONE  AND   CYLINDER.  21 1 

Ex.  593.  A  right  cylinder  of  volume  50  has  a  circum- 
ference of  9;    find  lateral  area.  \ 

Ex.  594.  In  a  right  circular  cylinder  of  volume  8,  the 
lateral  area  equals  the  sum  of  the  bases;   find  altitude. 

Ex.  595.  If  in  three  cylinders  of  the  same  height  one 
radius  is  the  sum  of  the  other  two,  then  one  lateral  area 
is  the  sum  of  the  others,  but  contains  a  greater  volume. 

Ex.  596.  What  is  the  relation  between  the  volumes 
of  two  cyliders  when  the  radius  of  one  equals  the  alti- 
tude of  the  other? 


CHAPTER  XV. 


PURE    SPHERICS. 


463.  If,  instead  of  the  plane  and  straight,  we 
take  the  sphere  and  its  great  circle,  that  is,  its 
geodesic  or  straightest,  then  much  of  our  plane 
geometry  holds  good  as  spherics,  and  can  be  read 
off  as  spherics.  Deducing  spherics  from  a  set  of 
assumptions  which  give  no  parallels,  no  similar  fig- 
ures, we  get  a  two-dimensional  non-Euclidean  geome- 
try, yet  one  whose  results  are  also  part  of  three- 
dimensional  Euclidean. 


I.  Assumptions  of  association  on  the  sphere. 

I  i\  For  every  point  of  the  sphere  there  is 
always  one  and  only  one 
other  point  which  with  the 
first  does  not  determine  a 
straightest.  This  second  point 
we  will  call  the  opposite  of  the 
first. 

Two   points,    not    each    the 
other's  opposite,  always  deter- 
mine a  straightest. 
Such  points  are  said  to  be  on  or  of  the  straightest, 
and  the  straightest  is  said  to  be  through  them. 

I  2'.  Every  straightest  through  a  point  is  also 
through  its  opposite. 


Fig.  170. 


PURE  SPHERICS. 

1 3'.  Any  two  points  of  a 
straightest,  not  each  the 
other's  opposite,  determine 
this  straightest;  and  on  every 
straightest  there  are  at  least 
two  points  not  opposites. 

1 4'.  There  are  at  least 
three  points  not  on  the  same 
straightest. 

464.  Theorem.     If  Of  is  the  opposite  of  0, 
O  is  the  opposite  of  0'. 

Proof.  If  0  is  not  the  opposite  of  0',  they  deter- 
mine a  straightest.  There  is  a  point  P  not  on  this 
straightest  (by  1 4'),  and  this  point  is  not  the 
opposite  of  0,  since  it  is  not  0'.  .*.  0,  P  deter- 
mine (by  Ii')  a  straightest  which  (by  I  2')  goes 
through  0' .     .'.  O,  Of  do  not  determine  a  straightest. 

465.  Theorem.  Two  distinct  straightests  can- 
not have  three  points  in  common.  [Proved  as  in 
6J 

II.  Assumptions  of  betweenness  on  the  sphere. 

466.  These  assumptions  specify  how  "  between" 
may  be  used  of  points  in  a  straightest  on  a  sphere. 

II  1'.  No  point  is  between  two  opposites. 

II  2'.  No  point  is  between  its  opposite  and  any 
third  point. 

II 3'.  Between  any  two  points  not  opposites 
there  is  always  a  third  point. 

II  4'.  If  B  is  between  A  and  C,  then  B  is  also 
between  C  and  A,  and  is  neither  C  nor  A. 

II  5'.  If  A  and  B  are  not  opposites,  then  there 


214 


RATIONAL   GEOMETRY. 


Fig.  172. 


is  always  a  point  C  such  that 
B  is  between  A  and  C. 

II  6r.  Of  any  three  points, 
not  more  than  one  can  be 
between  the  other  two. 

II  7'.  If  B  is  between  A 
and  C,  and  C  is  between  A 
and  D,  then  B  is  between  ^4 
and  D. 

II 8'.  Between  no  two  points  are  there  two 
opposites.  ■ 

467.  Definition.  Two  points  A  and  B,  not  oppo- 
sites, upon  a  straight  est  a,  we  call  a  s£c*  and  desig- 
nate it  with  AB  or  BA.  The  points  between  A 
and  B  are  said  to  be  points  of  the  sect  A  B  or 
also  situated  within  the  sect  A 5.  The  remaining 
points  of  the  straightest  a  are  said  to  be  situated 
without  the  sect  AB.  The  points  A,  B  are  called 
end-points  of  the  sect  AB. 

II  9'.  (Pasch's  assumption.) 
let  A,  B,  C  be  three  points, 
not  all  on  a  straightest,  and 
no  two  opposites,  and  let  a  be  a 
straightest  on  which  are  none 
of  the  points  A,  B,  C\  if  then 
the  straightest  a  goes  through 
a  point  within  the  sect  AB, 
it  must  always  go  either 
through  a  point  of  the  sect 
BC  or  through  a  point  of  the  sect  AC;  but  it  can- 
not go  through  both. 

468.  Theorem.     Every    straightest    a    separates 


On     the     sphere, 


Fig.   173. 


PURE  SPHERICS. 


215 


the  other  points  of  the  sphere  into  two  regions, 
of  the  following  character: 
every  point  A  of  the  one  re- 
gion determines  with  every 
point  B  of  the  other  region, 
not  its  opposite,  a  sect  AB 
within  which  lies  a  point  of  the 
straightest  a;  on  the  contrary, 

anv  two  points  A  and  A'  of 

;        4   ft  -1  FlG-  x74- 

one  and  the  same  region  al- 
ways determine  a  sect  A  A'  which  contains  no  point 
of  a. 

[Proved  as  in  22.] 

Points  in  the  same  one  of  these  two  regions  are 
said  to  be  on  the  same  side  of  a. 

469.  Theorem.  The  points  of  a  straightest  a 
other  than  two  opposites,  0,  0',  are  separated  by 
0,  0'  into  two  classes  such  that  0  or  0'  is  between 
any  point  of  the  one  and  any  non-opposite  point 
of  the  other,  but  neither  0  nor  0f  is  between  two 
of  the  same  class. 

Proof.  Take  any  other  straightest  b  through 
0  and  .'.  through  0'.  It  (by 
468)  cuts  the  sphere  into  two 
regions.  Now  (by  II  5 0  a  is 
not  wholly  in  either  of  these 
regions;  but  all  its  points 
other  than  0  and  0'  are  in 
these  regions.  Two  in  the 
same  region  have  no  point 
of  b  between  them.  But  0 
and  01  are  points  of  b.     Two 


Fig.  175. 


21 6  RATIONAL   GEOMETRY. 

not  opposites  in  different  regions  have  a  point  of 
b  between  them;    ,\  either  0  or  0'. 

470.  Definition.  The  parts  of  a  straightest 
determined  by  a  point  of  it  0  (with  its  opposite  0') 
are  called  rays  from  0. 

0  and  0'  are  called  end-points  of  the  rays. 
II  10'.  If  C  is  a  point  of  ray  PP\  every  other 
point  of  the  ray  is  between  C  and  P  or  C  and  P' . 

471.  Theorem.  Two  opposites  cannot  both  be 
on  the  same  ray. 

Proof.     II  3',  II  10'  and  II  2'. 

472.  Theorem.  Every  straightest  has  a  point 
in  common  with  any  other. 

Proof.  If  not,  consider  the  straightest  deter- 
mined by  any  point  of  the  one  and  a  point  of  the 
other.  This  would  have  on  one  ray  a  pair  of 
opposites,  contrary  to  471. 

473.  Definition.  On  the  sphere,  a  system  of 
sects,  A  B,  BC,  CD,  .  .  .  KL  is  called  a  sect- 
train,  which  joins  the  points  A  and  L  with  one 
another. 

The  points  within  these  sects  together  with 
their  end-points  are  all  to- 
gether called  the  points  o) 
the  sect-train. 

In  particular,  if  the  point 
L  is  identical  with  the  point 
A ,  then  the  sect-train  is  called 
a  spherical  polygon.  The  sects 
are  called  the  sides  of  the 
FlG-  I76-  spherical  polygon;    their  end- 

points  its  vertices. 


PURE  SPHERICS.  217 

Polygons  with  three  vertices  are  called  spherical 
triangles. 


Fig.  177.  Fig.  178. 

474.  Theorem.  Every  spherical  triangle  sepa- 
rates the  points  of  the  sphere  not  pertaining  to 
its  sect-train  into  two  regions,  an  inner  and  an 
outer.     [As  in  29.] 

475.  Convention.  On  a  given  straightest  OA, 
the  two  rays  00\  from  0  to  its  opposite  0',  are 
distinguished  as  of  opposite  sense.  This  distinc- 
tion may  be  indicated  by  a  qualitative  use  of  the 
signs  +  and  -  (plus  and  minus),  as  in  writing 
positive  and  negative  numbers. 

Any  sect  PO'  or  ray  from  P  through  0',  or  any 
sect  PB  where  B  is  between  P  and  0',  has  the  sense 
of  that  ray  00'  on  which  is  P. 

Then  also  BP  is  of  sense  opposite  that  of  PB. 

III.  Assumptions  of  congruence  on  the  sphere. 

Ill  1'.  If  A,  B  are  two  points,  not  opposite, 
on  a  straightest  a,  and  A'  a  point  on  the  same  or 
another  straightest  a',  then  we  can  find  on  a  given 
ray  of  the  straightest  ar  from  A'  always  one  and 


2l8 


RATIONAL    GEOMETRY. 


only  one  point  B' ,  such  that  the  sect  A B  is  con- 
gruent to  the  sect  A'Bf . 
Always  AB  =  AB  =  BA. 

Ill  2'.  If  AB  =  A'Bf  and  ,4B  =  A"£",  then  is 
also  A'B'^A"B". 

Ill  3r.  On  the  straightest  a  let  AB  and  BC  be 
two  sects  without  common  points,  and  further- 
more A'B'  and  B'C  two  sects  on  the  same  or 
another  straightest,  likewise  without  common 
points;  if  then  AB  =  A'B'  and  BC  =  B'C\  then 
is  also  AC^A'C. 

476.  Definition.     On  the  sphere,  let  h,  k  be  any 
two  distinct  rays  from  a  point  0,  which  pertain 
to  different  straightests.    These 
two  rays  h,  k  from  0  we  call 
a   spherical   angle,    and   desig- 
nate it  by  -4.  (h,  k)  or  ^  (&,  h). 
The  rays  /&,  &,  together  with 
the  point  0  separate  the  other 
points  of  the  sphere  into  two  re- 
gions, the  interior  of  the  angle 
Fig.  179.  anc[  the  exterior.     [As  in  35.] 

The  rays  h,  k  are  called 
sides  of  the  angle,  and  the 
point  0  the  vertex. 

Ill  4'.  On  the  sphere,  given 
a  spherical  angle  ^  (h,  k),  and 
a  straightest  a',  also  a  de- 
termined side  of  a'.  Designate 
by  h'  a  ray  of  the  straightest 
a'  starting  from  the  point  0' ': 
then  is  there  one  and  only  one  ray  kf  such  that  the 


Fig.  180. 


PURE  SPHERICS. 


219 


%.  (h,  k)  is     congruent     to     the     angle    ?£(h\    k'), 
and    likewise    all     inner     points     of      the     angle 
^  (h\  k')  lie  on  the  given  side  of  a! . 
Always  ?  (h,  k)  =  ^.  (h,  A?)  =  ?  (k,  h). 
Ill  5'.  If  ?  (h,  k)=$  {W,  k')   and    ?  (h,    fc)s 
?  (&",  &"),  then  is  also  *  (/*',  £')  ^  ?  (/*",  fe"). 

477.  Convention.  On  the  sphere  let  ABC  be 
an  assigned  spherical  triangle; 
we  designate  the  twro  rays 
going  out  from  A  through 
B  and  C  by  h  and  k  respect- 
ively. Then  the  angle  ^  (ht  k) 
is  called  the  angle  of  the 
triangle  ABC  included  by 
the  sides  AB  and  AC,  or 
opposite  the  side  BC. 

It  contains  in  its  interior  all  the  inner  points 
of  the  spherical  triangle  ABC 
and  is  designated  by  ^.BAC 
or  £A. 

Ill  6'.  On  the  sphere,  if  for 
two  triangles  ABC  and  A'B'C 
we  have  the  congruences 
AB^A'B',  AC^A'C,  ^BAC 
=  ^.B'AfC>  then  also  always 
are  fulfilled  the  congruences 
t  ABC ~t  A'B'C   and    ?ACB^  *A'CB'. 

478.  Convention.  When  the  sect  AB  is  set  off 
on  a  ray  starting  from  A,  if  the  point  B  falls 
within  the  sect  AC,  then  the  sect  A B  is  said  to  be 
less  than  the  sect  AC. 


Fig.   182, 


220 


RATIONAL   GEOMETRY. 


Fig.  183. 


In  symbols,  AB<AC. 

Then  also  AC  is  said  to  be 
greater  than  AB. 

In  symbols,  AC>AB. 

AB>  CD  when  E  between  A 
and  B  gives  A  is  =  CD  or  5£  = 
CD,  using  =  f or  m . 


479.  Convention.  When 
4-AOB  is  set  off  from 
vertex  0'  against  one  of  the 
rays  of  'ifA'O'C  toward  the 
other  ray,  if  its  second  side 
falls  within  i$A'&G%  then  the 
^AOB  is  said  to  be  less  than 
the  ^A'O'C. 

In  symbols, 

^-AOBK^A'O'C. 


Fig.  184. 


Then  also    ^A'O'C  is   said   to  be  greater  than 
4A0B. 
In  symbols,  ^.A'O'Cy^AOB. 


480.  Definition.  Two  spher- 
ical angles,  which  have  the 
vertex  and  one  side  in  common 
and  whose  not-common  sides 
make  a  straightest  are  called 
adjacent  angles. 


Fig.  185. 


PURE  SPHERICS, 


221 


481.  Definition.  Two  spher- 
ical angles  with  a  common 
vertex  and  whose  sides  make 
two  straightests  are  called  verti- 
cal angles. 


Fig.  186. 


Fig.  187. 


482.  Definition.  A  spherical 
angle  which  is  congruent  to 
one  of  its  adjacent  angles  is 
called  a  right  angle. 

Two  straightests  which 
make  a  right  angle  are  said 
to  be  perpendicular  to  one 
another. 


483.  Convention.  If  A,  B  are  points  which  deter- 
mine a  straightest,  then  we  may  designate  one  of 
the  regions  or  hemispheres  it  makes  as  right  from 
the  straightest  AB  taken  in  the  sense  of  the  sect 
AB  (and  the  same  hemisphere  as  left  from  BA 
taken  in  the  sense  from  B  to  A). 

If  now  C  is  any  point  in  the  right  hemisphere 
from  AB,  then  we  designate  that  hemisphere  of 
AC  in  which  B  lies  as  the  left  hemisphere  of  AC. 
So  we  can  finally  fix  for  each  straightest  which 
hemisphere  is  right  from  this  straightest  taken  in 
a  given  sense. 

Of  the  sides  of  any  angle,  that  is  designated  as 
the  right  which  lies  on  the   right   hemisphere  of 


222 


RATIONAL   GEOMETRY. 


that  straightest  which  is  determined  (also  in  sense) 
by  the  other  side,  while  the  left  side  is  that  lying 
on  the  left  of  the  straightest  which  is  determined 
(also  in  sense)  by  the  other  side. 

Two  spherical  triangles  with  all  their  sides  and 
angles  respectively  congruent  are  called  congruent 
if  the  right  side  of  one  angle  is  congruent  to  the 
right  side  of  the  congruent  angle,  and  its  left  side 
to  that  angle's  left;  but  if  the  right  side  of  one 
angle  be  congruent  to  the  left  side  of  the  con- 
gruent angle,  and  its  left  side  to  that  angle's  right, 
the  triangles  are  called  symmetric. 


484.  Theorem.  Two  spheri- 
cal triangles  are  either  con- 
gruent or  symmetric  if  they 
have  two  sides  and  the  in- 
cluded angle  congruent. 

[Proved  as  in  43.] 


Fig.  188. 


485.  Theorem.  Two  spheri- 
cal triangles  are  either  con- 
gruent or  symmetric  if  a  side 
and  the  two  adjoining  angles 
are  respectively  congruent. 

[Proved  as  in  44.] 


Fig.  189. 


PURE  SPHERICS. 


223 


486.  Theorem.     If  two  spherical  angies  are  con- 
gruent, so  are  also  their  adjacent  angles. 

[Proved  as  in  45.] 

487.  Theorem.       Vertical    spherical    angles    are 
congruent. 

[Proved  as  in  46.] 


488.  Theorem.     All  right  an- 
gles are  congruent. 
[Proved  as  in  50.] 


Fig.  190. 

489.  Theorem.  At  a  point  A  of  a  straightest  a 
there  is  not  more  than  one  perpendicular  to  a. 

[Proved  as  in  52.] 

490.  Definition.  When  any  two  spherical  angles 
are  congruent  to  two  adjacent  spherical  angles 
each  is  said  to  be  the  supplement  of  the  other. 

491.  Definition.  If  a  spher- 
ical angle  can  be  set  off  against 
one  of  the  rays  of  a  right  angle 
so  that  its  second  side  lies 
within  the  right  angle,  it  is 
called  an  acute  angle. 

Fig.  191. 
492.  Definition.      A  spherical  angle  neither  right 
nor  acute  is  called  an  obtuse  angle. 


224 


RATIONAL   GEOMETRY. 


493.  Definition.  A  spheri- 
cal triangle  with  two  sides 
congruent  is  called  isosceles. 


Fig.  192. 
494.  Theorem. 


The  angles  opposite  the  congru- 
ent sides  of  an  isosceles  triangle  are  congruent. 
[Proved  as  in  57.] 

495.  Theorem.  If  two  angles  of  a  spherical  tri- 
angle be  congruent,  it  is  isosceles. 

[Proved  as  in  485.] 


496.  Theorem.  Two  spher- 
ical triangles  are  either  con- 
gruent or  symmetric  if  the 
three  sides  of  the  one  are  con- 
gruent, respectively,  to  the 
three  sides  of  the  other. 

[Proved  as  in  58.] 


Fig.  193. 


497.  Theorem.         Any   two 

straightests  perpendicular  to  a 
given  straightest  intersect  in  a 
point  from  which  all  sects  to  the 
given  straightest  are  perpendic- 
ular to  it  and  congruent. 


Fig.  194. 


PURE  SPHERICS. 


225 


Given  r't  ^A=  ifC. 

To  prove  PA=PC=PD, 
and   ^Drt. 

Proof.     By  495,  PA=    pC 

and     (by     485)     PA^P'A. 

;.  (by  484)  4PDA=  ?P'DA; 

.'.  (by  488)  %PDA=  ^.PAD; 

.'.(by  495)         PDmPA. 

Fig.  195. 

498.  Definition.  The  two  opposite  points  at  which 
two  perpendiculars  to  a  given  straightest  intersect 
are  called  its  poles,  and  it  the  polar  of  either  pole. 

A  sect  from  a  pole  to  its  polar  is  called  a  quadrant. 

499.  Theorem.       All  quadrants  are  congruent. 
Let  AB  and  A'B'  be  two 

quadrants. 

To  prove  AB  =  A'Bf. 
Proof.      At  A  take  a  r't? 
£AC,  and  also  at  A'.    On  AC 
take  a  sect  AC,  and  on  AC 
take  A'C7  a  AC.     At  C  and  C 
Fig.  196.  take    straightests    ±AC    and 

ArCr .    These  contain  B  and 
B'.     :.  (by  485)  AB  =  A'B'. 

500.  Theorem.  A  ^oiw/ 
which  is  a  quadrant  from  two 
points  of  a  straightest  not  oppo- 
sites  is  its  pole. 

Let  PA  and  PC  be  two 
quadrants. 

Proof,      At  A  and  C  erect 


226 


RATIONAL   GEOMETRY. 


perpendiculars  intersecting  at  P' .      Then  (by  499 
and  496)    IPAC^P'AC. 

501.  Theorem.  //  three  sects  from  a  point  to  a 
straightest  be  equal,  they  are  quadrants. 

Proof.  They  are  sides  of  two  adjacent  isosceles 
triangles,  and  hence  perpendiculars. 

502.  Contranominal  of  501.  If  three  equal  sects 
from  a  point  be  not  quadrants,  their  three  other  end- 
points  are  not  on  a  straightest. 


503.    Theorem. 
straightest  a, 


Through  a  point  A,   no'    on  a 
there  is  to  a  always  a  perpendicular 
straightest  which,  if  A  be  not 
a  pole  of  a,  is  unique. 

Proof.  Take  any  sect  QR 
on  a.  Take  on  the  other  side 
of  a  from  A,  ?BQR=  4-AQR, 
and  QB  =  QA. 

Then  (by  484)  AB±  a  at 
5. 

Moreover,   if  there  were  a 
second  straightest  perpendic- 
ular to  a  from  A,  then  A  would  (by  498)  be  a  pole 
of  a. 

504.  Definition.  A  point  B  of  a  given  ray  00' 
such  that  BO  s  BO'  will  be  called  the  bisection-point 
of  the  ray.  A  point  B  between  A  and  C  such  that 
AB  =  BC  is  called  the  bisection-point  of  the  sect 
AC. 

505.  Problem.  To  bisect  a  given  ray 
00'. 

Construction.    At  two  points  of  the  given  ray  not 


PURE  SPHERICS. 


227 


both  end-points  erect  perpen- 
diculars [take  (by  III  4')  an- 
gles =  to  ^5  in  503],  inter- 
secting at  P.  Take  another 
ray  from  0,  not  on  the  same 
straightest  as  the  given  ray, 
and  at  two  points  of  it  not 
both  end-points  erect  perpen- 
diculars intersecting  at  Q. 
The  straightest  PQ  bisects  the 
given  ray  00'. 

Proof.  Since  P  and  Q  are  poles,  .'.  ^.B  = 
r't  ?=?£>.     A  (by  485)  OB  =  BO'. 

506.  Theorem.  If  0  and  0'  are  opposites,  then 
with  vertex  0  ^  (h,  k)  =  ^  (h,  k)  with  vertex  0' . 

Proof.  Bisect  (by  505)  ray  h  at  A  and  ray  k  at  C. 
Then  (by  496)   %AOC=  %AO'C. 

507.  Definition.  From  the  vertex  0,  a  ray  b  with- 
in ?(/&,  k)  making  ^  (h,  b)=  £(&,  k)  will  be  called 
the  bisector  of  $  (h,  k). 

508.  Problem.  To  bisect  a  given  spherical  angle. 
Construction.  By  505,  bi- 
sect the  rays  of  the  angle 
?B  at  H  and  F.  Take  A 
between  H  and  B  and  from 
F  on  FB'  take  FC^HA. 
Then  ^C  intersects  HF  at  Z), 
and  £D  bisects  ^HBF. 

Proof.  By  496,  %ACB= 
? CAS';. -.by 485,  HD~FD\ 
.\  by  496,     *  tf££  =  ?  F££>. 


228 


RATIONAL   GEOMETRY, 


509.  Problem.  To  bisect  a 
given  sect. 

Construction.  At  the  end- 
points  erect  perpendiculars 
[by  taking  (by  III  4')  angles 
=  ?S  in  503].  ^ 

Bisect  (by  508)  the  £  be- 
tween them. 

Proof.     By  497  and  484.  FlG-  2DI- 

510.  Corollary.  In  an  isosceles  triangle  the  bi- 
sector of  the  angle  between  equal  sides  bisects  at 
right  angles  the  third  side. 

5 xx.  Theorem.  If  two  spher- 
ical triangles  have  two  sides 
of  the  one  equal  respectively 
to  two  sides  of  the  other,  and 
the  angles  opposite  one  pair  of 
equal  sides  equal,  then  the 
angles  opposite  the  other  pair 
are  either  equal  or  supple- 
Fig.  202.  mental. 

[Proved  as  in  175.] 

512.  Definition.  In  any  spherical  triangle  the 
sect  having  as  end-points  a  vertex  and  the  bisec- 
tion-point of  the  opposite  side  is  called  a  median. 

513*  Theorem.  An  angle  adjacent  to  an  angle  of 
a  spherical  triangle  is  greater  than,  equal  to,  or  less 
than  either  of  the  interior  non-adjacent  angles,  accord- 
ing as  the  median  from  the  other  interior  non-adjacent 
angle  is  less  than,  equal  to,  or  greater  than  a  quadrant. 
And  inversely. 
Proof.    Let  ^  ACD  be  an  angle  adjacent  to  ^ACB 


PURE  SPHERICS. 


229 


Fig.  203. 


of  aACB.  Bisect  AC  at  F. 
On  straightest  BF  beyond  F 
take  FH  =  FB.  .\  (by  484) 
^BAF=  ^HCF.  If  now  the 
median  BF  be  a  quadrant 
BFH  is  a  ray  and  H  is  on 
£CX>.  If  the  median  BF  be 
less  than  a  quadrant,  if'  is 
within  IfACD. 

.-.  ^/fCA^JDCA  .-.  jDCA  >  ?£AC. 

If  BC  be  greater  than  a  quadrant,  if"  is  without 

.-.  1H"CF>*DCF.  .\  ?£>CV1<?£,4C\ 

514.  Definition.  Two  sects  respectively  congru- 
ent to  two  made  by  a  point  on  a  ray  with  its  end- 
points  are  called  supplemental. 

515.  Theorem.  The  supplements  of  congruent 
sects  are  congruent. 

Proof.  They  are  sums  or  differences  of  quadrants 
and  congruent  sects  less  than  quadrants;  and  (by 
499)  all  quadrants  are  congruent. 

516.  Theorem.  If  a  median 
be  a  quadrant,  it  is  an  angle- 
bisector,  and  the  sides  of  the 
bisected  angle  are  supplemental. 
Let  median  BD  in  a  ABC 
be  a  quadrant. 

Proof.  In  2ABD  and 
aCB'D  (by  484)  AB  = 
CBf  and  ^ABD  =  ^CB'D  = 
^CBD  (by  506). 


Fig. 


204. 


23° 


RATIONAL   GEOMETRY. 


Given  ^.B=^E; 


517.  Inverse.  If  two  sides  of  a  triangle  are  sup- 
plemental, the  median  is  a  quadrant. 

Proof.     2  A  BC  =  or  -V  2AB'C  (by  496). 
.-.  2ABD  =  or  -|.  2CB'D  (by  484).     ;'-  BD=DB'. 

518.  Corollary.  If  two  sides  of  a  triangle  are  sup- 
plemental, the  opposite  angles  are  supplemental. 

519.  Theorem.  Two  spherical  triangles  are  either 
congruent  or  symmetric  if  they  have  two  angles  of  the 
one  respectively  equal  to  two  of  the  other,  the  sides  oppo- 
site one  pair  equal,  and  those  opposite  the  other  pair 
not  supplemental. 

$C=^F;AB  =  DE.;  AC  not 
supplemental  to  FD. 

Proof.  On  ray  BC  take 
BG  =  EF.  G  must  be  C,  else 
would  we  have  a  'aACG  with 
adjacent  ^-AGB  =  ?{ACG  in- 
terior non-adjacent  and  .*. 
with  median  a  quadrant  (by 
513)  and  .-.  (by  516)  with  AC 
supplemental  to  AG,  that  is, 
to  FD. 

520.  Theorem.  Two  spheri- 
cal triangles  are  either  congru- 
ent or  symmetric  if  they  have 
in  each  one,  and  only  one, 
right  angle,  equal  hypothenuses 
and  another  side  or  angle  con- 
gruent. 

Given  ifCm^Hmfi  ?, 
and  c  =  h.  If  a=f,  then  if 
AC>g,takeCD=g.     .\BD  = 


Fig.  205. 


Fig.  206. 


h  =  c,  and  (by  510)  the 


PURE  SPHERICS.  231 

bisector  of  $DBA  is  _L  to  CD  A.     .*.  ^by  498)  B  is 
pole  to  CD  A.     .'.  %-A  is  also  r't. 

If  %.A  m  ?F,  then  if  $ABC  >?£,  take  $ABD  = 
*G.  .'.  (by  485)  ^BDA  =  ?tf  =  ?C  =  r't  ?.  .-.  £ 
is  pole  toCZM. 

521.  Theorem.  The  straight- 
est through  the  poles  of  two 
straightests  is  the  polar  of  their 
intersection-points. 

Let  A  and  B  be  poles  of  a 

and  b,  which  intersect  in  P. 

To  prove  A  B  the  polar  of  P. 

Proof.     AP    and    £P    are 

,  Fig.  207. 

quadrants. 

522.  Corollary  to  521.  The  straightest  through 
the  poles  of  two  straightests  is  perpendicular  to 
both. 

523.  Corollary  to  521.  If  three  straightests  are 
copunctal,  their  poles  are  on  a  straightest. 

524.  Definition.  If  A,  B}  C  are  the  vertices  and 
a,  b,  c  the  opposite  sides  of  a  spherical  triangle,  and 
Af  that  pole  of  a  on  the  same  side  of  a  as  A,  B'  of 
b  as  By  C  of  c  as  C,  then  A'B'C  is  called  the  polar 
triangle  of  ABC. 

525.  Definition.  Of  a  spherical  triangle  A}  B,  C, 
the  polar  triangle  is  A' ',  B',  C  where  A'  is  that  pole 
of  BC  or  a  on  the  same  side  of  a  as  A,  Br  of  b  as  £, 
C  of  c  as  C 

526.  Theorem.  If  of  two  spherical  triangles  the 
second  is  the  polar  of  the  first,  then  the  first  is  the  polar 
of  the  second. 

Let  ABC  be  the  polar  of  A'B'C 


232 


RATIONAL    GEOMETRY. 


Fig.   208. 


To  prove  A'B'C  the  polar  of  ABC. 
Proof.  Since  B  is  pole  of  AfC,  .'.  BA'  is  a  quad- 
rant; and  since  C  is  pole  of 
A'B',  .' .  CA'  is  a  quadrant; 
•  '•  (by  500)  A'  is  pole  of  BC. 
In  like  manner,  B'  is  pole  of 
AC,  and  C  of  AB.  More- 
over, since  by  hypothesis  A 
and  A'  are  on  the  same  side 
of  B'C  and  A  is  pole  of  B'C, 
.'.  sect  A  A'  is  less  than  a  quad- 
rant. .'.  A  and  A1  are  on  the  same  side  of  BC,  of 
which  A'  is  pole.     And  so  for  B'  and  O \ 

527.  Theorem.  J«  a  /?<wr  0/  £<?/ar  triangles,  any 
angle  of  either  intercepts,  on  the  side  of  the  other  which 
lies  opposite  it,  a  sect  which  is  the  supplement  of  that 
side. 

Let  ABC  and  A'B'C  be 
two  polar  triangles. 

Proof.  Call  D  and  E  re- 
spectively the  points  where 
ray  A'B'  and  ray  A'C  meet 
BC.  Since  B  is  pole  of  A'C, 
.'.  BE  is  a  quadrant,  and  since 
C  is  pole  of  A'B',  .'.  CD  is  a 
quadrant. 


Fig.  209. 


But  BE  +  CD=BC  +  CE  +  CB  +  BD  = 

BC  +  (£C  +  CB  +  £L>)  =  BC  +  £>£. 

528.  Theorem.  Two  spherical  triangles  are  either 
congruent  or  symmetric  if  they  have  three  angles  of 
the  one  respectively  equal  to  three  angles  of  the  other. 


PURE  SPHERICS. 


233 


Proof.  Since  the  given  triangles  are  respectively 
equiangular,  their  polars  are  respectively  equilateral. 
For  (by  484)  equal  angles  at  the  poles  of  straightests 
intercept  equal  sects  on  those  straightests;  and 
these  equal  sects  are  the  supplements  of  correspond- 
ing sides  of  the  polars  Hence  these  polars,  having 
three  sides  respectively  equal,  are  respectively 
equiangular.  Therefore  the  original  triangles 
are  respectively  equilateral,  which  was  to  be 
proved. 

529.  Corollary  to  511.  Two  spherical  triangles 
are  either  congruent  or  symmetric  if  they  have  two 
sides  of  the  one  respectively  equal  to  two  of  the 
other,  the  angles  opposite  one  pair  equal,  and  those 
opposite  the  other  pair  not  supplemental. 

530.  Theorem.  //  two  sides  of  a  spherical  triangle 
are  each  less  than  a  quadrant,  any  sect  from  the  third 
side  to  the  opposite  vertex  is  less  than  a  quadrant. 

Let  A  B  and  BC  be  each  less  than  a  quadrant. 

To  prove  BD  <  quadrant. 

Proof.  Let  FG,  the  polar  of 
B,  meet  BD  at  H.  If  H  were 
between  B  and  D,  then  GHF 
would  (by  II  9')  meet  CAt 
and  so  have  a  point  on  each 
of  the  three  sides  of  a  AB'C, 
which  (by  II  9' )  is  impossi- 
ble. Hence  D  is  between  B 
and  H.  That  is  BD<  quad- 
rant. 

531.  Corollary  to  530  and  513.  If  two  sides  of  a 
spherical  triangle  are  each  less  than  a  quadrant,  the 


Fig.  210. 


234 


RATIONAL    GEOMETRY. 


Fig.  211. 


angle  opposite  either  is  less  than  the  supplement  of 
the  angle  opposite  the  other. 

532.  Theorem.  If  two  sides  of  a  spherical  triangle 
be  each  less  than  a  quadrant,  as  the  third  side  is  greater 
or  less  than  one  of  these,  so  is  it  with  the  opposite  angles. 
And  inversely. 

Let  in  2 ABC,  BC  and 
another  side,  AB,  be  each 
less  than  a  quadrant,  and 
AC>AB. 

To  prove  ^fABC>^ACB. 

Proof.  Within  AC  take  D 
making  AD  =  AB.  Then  (by 
530)  DB  is  less  than  a  quad- 
rant. .'.  (by  531)  1ADB> 
?  C.     But  2 ABC  >  ^ABD  m  %ADB  >  ?C. 

533.  Theorem.  If  the  three  sides  of  a  spherical  tri- 
angle are  each  less  than  a  quadrant,  any  two  are 
together  greater  than  the  third. 

[Proved  as  in  174.] 

534.  Definition.  On  the  sphere,  the  assemblage 
of  points  which  with  a  given  point  give  congruent 
sects  is  called  a  circle.  The  given  point  is  called  a 
pole  of  the  circle.  Any  one  of  the  congruent  sects 
is  called  a  spherical  radius  of  the  circle. 

Thus  a  straightest  is  a  circle  with  a  quadrant  for 
spherical  radius.  But  henceforth,  for  convenience, 
by  circle  we  will  mean  a  circle  with  a  radius  not  a 
quadrant. 

A  sect  whose  end-points  are  on  a  circle  is  called 
a  spherical  chord,  or  simply  a  chord. 

A  chord  containing  a  pole  is  called  a  diameter. 

Since  the  supplements  of  congruent  sects  are  (by 


PURE  SPHERICS. 


235 


Fig.  212. 


515)  congruent,  therefore  every  circle  has  two  poles 
which  are  opposite  points,  and  its  spherical  radius 
to  one  pole  is  the  supplement  of  that  to  the  other. 

Always  one  spherical  radius  is  less  than  a  quadrant. 

Call  its  pole  the  g-pole,  and  it  the  ^-radius. 

535.  Theorem.  Any  spheri- 
cal chord  is  bisected  by  the 
perpendicular  from  a  pole. 

Proof.     AD=BD  (by  520). 

536.  Corollary.  A  straight- 
est  perpendicular  to  a  diam- 
eter at  an  end-point  has  only 
this  point  in  common  with 
the  circle. 

537.  Definition.  A  straightest  with  one  and  only 
one  point  in  common  with  a  circle  is  called  a  tangent 
to  the  circle. 

538.  Theorem.  //  an  oblique  from  a  point  to  a 
straightest  be  less  than  a  quadrant,  then  there  is  one 
and  only  one  perpendicular  sect  from  the  point  to  the 
straightest  which  meets  it  at  less  than  a  quadrant  from 
the  foot  of  the  oblique  and  this  is  less  than  a  quadrant. 

Let  BA  be  oblique  to  CA 
and  <q,  and  BC  J_CA. 

Proof.  Then  CA  cannot 
=  q,  else  would  BA  =  q.  Hence 
CA  may  be  taken  <q,  since 
from  C  to  its  opposite  =  2 q. 
Now  take  CAf  =CA.  Then 
BA' =BA  and  BC  is  median 
where  the  two  sides  are  each 
<q.  .;.  (by  530)  BC<q.  :.  (by  503)  its  prolonga- 
tion BC  is  the  only  other  1_  from  B  to  AC. 


Fig.  213. 


236 


RATIONAL    GEOMETRY. 


Fig.  214. 


539.  Definition.  If  A  be  a  point  of  a  circle  whose 
g-pole  is  P,  then  P  or  any  point  between  A  and  P  is 
said  to  be  within  the  circle,  while  Q  such  that  A  is 
between  P  and  Q  is  said  to  be  without  the  circle. 

540.  Theorem.  Any  straightest  through  an  end- 
point  of  a  diameter,  but  not  per- 
pendicular to  the  diameter,  has 
a  point  within  and  a  second 
point  on  the  circle. 

Let  P  be  the  g-pole,  and  AC 
the  straightest  through  A,  an 
end-point    of  diameter   BPA 
(i-PAC  not  r't). 
Proof.   Take  (  538)  PD±CA 
with  PD  and  AD  each  < q.     Take  DC  =  DA. 

.  • .  (by  484)  PC  =  PA\  that  is,  C  on  circle.  More- 
over (by  5 1 3)  2PAF  >  ^PCA  =  2PAC.  .  \  %PAD 
acute,  /.(by  532)  PD<PA;  that  is,  D  within 
circle. 

541.  Theorem.  Any  straightest  with  a  point  on 
and  a  point  within  a  circle  has  a  second  point  on  the 
circle. 

Let  A  B  have  a  point  A  on 
and  B  within  circle  with  q- 
pole  P. 

Proof.  ifPAB  cannot  be 
r't.  For  if  so,  then  produc- 
ing PB  to  meet  the  circle  at 
C,  (by  531)  ^PCF>1PAC 
>r't  $PAB.  .«.  If  PC  A  ad- 
jacent to  obtuse  ^PCF  is  FlG-  2I5- 
acute.      But    it    is    also    obtuse,    being    (by    494) 


PURE  SPHERICS 


237 


=  ifPAC.      This  is    impossible,    .-.  BA   not  A_AP\ 
.'.  (by  540)  it  has  a  second  point  on  the  circle. 

542.  Corollary.  A  tangent  has  no  point  within 
the  circle. 

543.  Theorem.      //  less  than  a  quadrant,  the  per- 
pendicular is  the  least  sect  from 
a  point  to  a  straightest. 

Proof.  If  any  other  sect 
from  P  to  AC  were  less  than 
the  perpendicular  PA,  then 
AC  would  have  a  point  within 
the  circle  with  q-pole  P  and 
^-radius  PA,  and  .'.  (by  541) 
a  second  point  on  this  circle, 
which  (by  536)  is  impossible. 

544.  Convention.  In  general  a  sum  of  sects  is  a 
number  of  quadrants  plus  a  sect. 

545.  Theorem.  Any  two  sides  of  a  spherical  tri- 
angle are  together  greater  than  the  third. 

Proof.  Since  each  side  is  less  than  two  quadrants, 
we  have  only  to  prove  AB  +  BC>AC  when  AB<qt 
and  BC<  AC. 


Fig.  216. 


I.  If  BC  =  q,  then  taking 
CD=q,  we  have  (by  500) 
BD  i.  AC. 

.:<y>yS43)AD<AB. 

.-.    AC=AD+  DC  <AB  + 

BC. 


Fig.  217. 


238 


RATIONAL   GEOMETRY. 


II.  If  BC<q, 

(i)  if  CA<q,  this  is  533. 

(2)  if  C4  =  q,  erect  _L  at  A. 

(by    543)    AByBD. 

AB+BC>DB+BC=  DC 
-AC. 


Fig.  218. 

(3)  If  CA  >  <?,  take  on  it  CD  =  q  and  make  CBF  =  9. 
Then  sects  AB  and  Di7  cross  at 
K,  for  F  is  on  the  non-C-side 
of  A  B,  while  D  is  on  the  C-side 
of  AB.  .'.  DF  must  have  a 
point  on  straight  AB.  But 
all  points  of  sect  DF  are  inte- 
rior to  ^fC, .'.  this  intersection 
point  is  on  sect  A B,  which  is 
all  of  straight  AB  within  i£C. 

Then  (by  543)  BF<BK  and  AD<AK. 

.'.  AC  =  AD  +  DC<AK  +  CD 

=  AK  +  CF<CB  +  BK  +  KA. 

III.  If  BC>q}  then  in  ABC  all  sides  are  less  than 
quadrants. 

•*•  (by  533)  CB  +  BA  +  AC  >CB  +  BC  =CA  +  AC. 
.'.CB  +  BA>CA. 

546.  Definition.  A  convex  spherical  polygon  is 
one  no  points  of  which  are  on  different  sides  of  the 
straightest  of  any  of  its  sides. 

547.  Theorem.  A  convex  spherical  polygon  is  less 
than  one  containing  it. 


PURE  SPHERICS. 


239 


548.  Theorem.     The  sum  of  the  sides  of  a  convex 
spherical  polygon  is  less  than  four  quadrants. 

Proof.  It  is  within,  hence  less  than,  any  one  of 
its  angles. 

549.  Theorem.  If  one  angle 
of  a  spherical  triangle  be  greater 
than  a  second,  the  side  opposite 
the  first  must  be  greater  than  the 
side  opposite  the  second ;  and 
inversely. 

Given  ?C>  ?£. 

Proof.  Take  %DCB  =  ?£. 
Then(by495)ZX7  =  L>J3.  But 
(by  545)  DC  +  DA  >  AC. 

550.  Theorem.  In  a  cyclic  quadrilateral,  the  sum  of 
one  pair  of  opposite  angles  equals  the  sum  of  the  other 
pair. 

Proof.     By  isosceles  triangles. 


551.  Theorem.  Of  sects  join- 
ing two  symmetrical  points  to  a 
third,  that  cutting  the  axis  is 
the  greater. 

Proof.        BA=BD  +  DA 

=  BD  +  DA'>BA'. 


Fig. 


Fig.  221, 


552.  Theorem.  //  two  spherical  triangles  have  two 
sides  of  the  one  equal  to  two  sides  of  the  other,  but  the 
included  angles  unequal,  then  that  third  side  is  the 
greater  which  is  opposite  the  greater  angle',  and  in- 
versely. 


240 


RATIONAL   GEOMETRY. 


Fig. 


Proof.  Against  one  of  the  equal  sides  of  one  tri- 
angle construct  a  triangle  with  elements  equal  to 
those  in  the  other.  Bisect  the  angle  made  by  the 
pair  of  equal  sides.  This  axis  cuts  the  third  side, 
which  is  opposite  the  greater  angle. 

553.  Theorem.  //  each  of  the  two  sides  about  a 
right  angle  is  less  than  a  quad- 
rant, then  the  hypothenuse  is 
less  than  a  quadrant. 

Proof.  Extend  the  two 
sides  BA,  BC,  taking  BF  = 
BD=  quadrant.  Then  (by 
500)  B  is  pole  of  DF.  .'.  (by 
498  and  497)  ^Fisr't.  .'.(by 
495)  DF  is  a  quadrant.  .  \  (by 
500)  DA  is  a  quadrant.     /.  (by  530)  AC < quadrant. 

554.  Inverse  of  553. 

If  the  hypothenuse  and  a  side  are  each  less  than 
a  quadrant,  then  the  other  side  is  less  than  a  quad- 
rant. 

Proof.  If  B  is  r't  (Fig.  222),  and  AB  and  AC 
each  <q,  there  is  (by  538)  on  st'  AB  a  p't  H  such 
that  CH  and  AH  each  <  q  while  CH  JL  AH. 

But  H  is  B  or  B'. 

It  cannot  be  B'  since  BA<q  and  .*.  (by  II  10') 
AB'yq. 

555.  Theorem.  The  straightest  bisecting  two  sides 
of  a  triangle  meets  the  third  side  at  a  quadrant  from  its 
bisection-point. 

Let  the  straightest  through  A\  B\  the  bisection- 
points  of  two  sides  BC,  CA,  meet  the  third  side  pro- 
duced at  D  and  D'. 


PURE  SPHERICS. 


241 


Fig.  223. 


Proof.     Take  (by  538)  AL,  BM,  CN  J_  A'B'  and 
such  that  each  is  <q,  and 
also  B'L,  B'N,  A'M,  A'N 
each  <q. 

.-.  in2's  ALB'andCNB' 
(by  519)  AL  =  CN.  Simi- 
larly BM=CN.  .*.  in  a's 
AL£>  and  BMD'  (by  519) 
AD=BD'.  /.if  C  be  bi- 
section-point of  AB,  we 
have  C,A+^L>=C,5  +  JBJD,=^. 

556.  Theorem.  The  end-points  of  any  sect  taken 
with  any  point  on  its  perpendicular  bisector  give 
equal  sects. 

557.  Corollary  556.  Every  point  on  the  perpen- 
dicular bisector  of  a  sect  is  pole  of  a  circle  through 
its  end-points. 

558.  Corollary  to  557. 

The  perpendicular  bisectors  of  the  sides  of  a 
spherical  triangle  are  copunctal  (in  its  circumcenter) . 

559.  Corollary  I  to  555. 

The  altitudes  of  a  spherical  triangle  are  copunctal 
(in  its  orthocenter. 

For,  regarding  A'B'C  as  the  triangle,  the  perpen- 
dicular to  DC  at  C  is  the  polar  of  D,  and  .'.  J_  to 
A'B'. 

Similarly,  the  perpendicular  to  BA'  at  A'  is  _L 
to  B'C,  etc. 

So  the  three  altitudes  of  A'B'C  are  copunctal  in 
the  circumcenter  of  ABC. 

560.  Corollary  II  to  555.     (Lexell). 

The  vertices  of  spherical  triangles  of  the  same 


242 


RATIONAL   GEOMETRY. 


angle-sum  on  the  same  base  are  on  a  circle  copolar 
with  the  straightest  bisecting  their  sides. 

For  AO=BO,  ^OAB  =$OBA,  ^.LAB  =  ^MBA 
=  i[A+B  +  C].  Hence  a  A  OB  is  fixed,  and  .'.  OC 
[supplemental  to  OA]. 

561.  Theorem.  The  straight- 
ests  through  the  corresponding 
vertices  of  a  triangle  and  its 
polar  are  copunctal  in  the  com- 
mon orthocenter. 

Proof.  For  AA'  is  J_  to 
BC  and  B'C\  since  it  passes 

through  their  poles. 
Fig.  224. 


Equivalence. 

562.  Theorem.  Any  angle  made  with  a  side  of  a 
spherical  triangle  by  joining  its 
end-point  to  the  circumcenter, 
equals  half  the  angle-sum  less 
the  opposite  angle  of  the  tri- 
angle. 

Proof.  For  ?  A  +  ?B  +  ?C 
-  2  £  OCA  +  2  ?  OCB  ± 
2^.0AB.  .'.  %OCA  =  \&A 
+  %B  +  $C]  -  [?  OCB  ± 
$OAB]  =  M  $A  +  ?  B  +  ?C]  -  ?£. 

563.  Corollary  to  562.     Symmetrical  spherical  tri- 
angles are  equivalent  or  equivalent  by  completion. 

For  the  three  pairs  of  isosceles  triangles  formed 
by  joining  the  vertices  to  the'  circumcenters,  hav- 


Fig.  225. 


PURE  SPHERICS. 


243 


Fig.  226. 


ing  respectively  a  side  and  two  adjoining  angles 
congruent,  are  congruent. 

564.  Theorem.  Of  the  triangles  formed  by  three 
non-copunctal  straightests,  two  containing  vertical 
angles  are  together  equivalent  to  that  angle. 

To  prove  2ABC  +  2AB'C 
=  ^.ABA'CA. 

Proof.  B'C=BC,  each 
being  supplement  of  CB'. 
Again  A  C  =  A  'C  (supple- 
ments of  AC).  Again  AB'  = 
A'B  (supplements  of  AB). 
.-.  (by  496)  2ABfC  =  2BCAf. 

.-.  2  ABC  +  2AB'C  =  a 
ABC+2BCA' ^ABA'CA. 

565.  The  spherical  excess,  e,  of  a  spherical  triangle 
is  the  excess  of  the  sum  of  its  angles  over  two  right 
angles. 

In  general  the  spherical  excess  of  a  spherical  poly- 
gon is  the  excess  of  the  sum  of  its  angles  over  twice 
as  many  right  angles  as  it  has  sides  less  two. 

566.  Theorem.    A  spherical  triangle  is  equivalent  to 
half  its  spherical  excess. 

Proof.  Produce  the  sides 
of  the  a  ABC  until  they  meet 
again  two  and  two  at  A',  Bfy 
C .  The  a  ABC  now  appears 
in  three  angles,  if  A,  ^.B,  %.C. 
But  (by  564)  $A  =  2aBC  + 
2AB'C.     .-.  ?A  +  ?£4  ?C 

FlGl  227-  =2  r't?s  +  2^5C. 

2AAJ3C  =  ?A  +  ?£+?C-2r't  ?'s=e. 


244  RATIONAL   GEOMETRY. 

567.  Corollary  I  to  566.  The  sum  of  the  angles 
of  a  a  is  >  2  r't  f's  and  <  6  r't  ?'s. 

568.  Corollary  II  to  566.     Every  ?  of  a  A  is  >\e. 

569.  Corollary  III  to  566.  A  spherical  polygon  is 
equivalent  to  half  its  spherical  excess. 

Ex.  597.  If  a  spherical  angle  adjacent  to  one  angle  of  a 
spherical  triangle  is  equal  to  a  second  angle  of  the 
triangle,  the  sides  opposite  these  are  together  a  ray. 

Ex.  598.  In  a  spherical  triangle  and  the  spherical 
triangle  determined  by  the  opposites  of  its  vertices  the 
sides  and  angles  are  respectively  congruent. 

Ex.  599.  Where  are  the  vertices  of  spherical  triangles 
on  a  given  base  the  sum  of  whose  other  sides  is  a  ray  ? 

Ex.  600.  Does  a  triangle  ever  coincide  with  its  polar? 

Ex.  601.  The  difference  of  any  two  angles  of  a  spherical 
triangle  cannot  exceed  the  supplement  of  the  third. 

Ex.  602.  The  bisector  of  an  angle  passes  through  the 
pole  of  the  bisector  of  the -supplemental  adjacent  angle. 

Ex.  603.  If  two  straigh tests  make  equal  angles  with 
a  third,  the  sects  from  their  poles  to  its  are  equal. 

Ex.  604.  If  a  straightest  be  through  the  pole  of  a  second, 
so  is  the  second  through  a  pole  of  the  first. 

Ex.  605.  If  two  circles  be  tangent,  the  point  of  contact 
is  on  their  center-straightest. 

Ex.  606.  The  common  secant  of  2  intersecting  0s  bisects 
a  common  tangent. 

Ex.  607.  The  three  common  secants  of  3  ©s  which 
intersect  each  other  are  copunctal. 

Ex.  608.  If  a  quad'  can  have  a  0  inscribed  in  it,  the 
sums  of  the  opposite  sides  are  = . 

Ex.  609.  If  two  equal  ©s  intersect,  each  contains  the 
orthocenters  of  as  inscribed  in  the  other  on  the  common 
chord  as  base. 

Ex.  610.  Three  equal  ©s  intersect  at  a  point  H,  their 
other  points    of    intersection  being  A,  B,  C.     Show  that 


PURE  SPHERICS.  245 

H  is  the  orthocenter  of  a  ABC;  and  that  the  a  formed 
by  the  centers  of   the   circles  is  =to  a  ABC. 

Ex.  611.  The  feet  of  ±s  from  A  of  a  ABC  on  the  ex- 
ternal and  internal  bi's  of  ^  s  B  and  C  are  co-st'  with 
the  bisection-points  of  b  and  c.     Does  this  hold  for  a  ? 

Ex.  612.  (Bordage.)  The  centroids  of  the  4  as  de- 
termined by  four  concyclic  points  are  concyclic. 

Ex.  613.  The  orthocenters  of  the  4  as  determined  by 
four  concyclic  points,  A,  B,  C,  D,  are  the  vertices  of  a  quad' 
=  to  ABCD.  The  incenters  are  vertices  of  an  equian- 
gular quad'. 

Ex.  614.  (Brahmegupta.)  If  the  diagonals  of  a  cyclic 
quad'  are  JL,  the  -L  from  their  cross  on  one  side  bisects 
the  opposite  side. 

Ex.  615.  If  the  diagonals  of  a  cyclic  quad'  are  1 ,  the 
feet  of  the  ±s  from  their  cross  on  the  sides  and  the  bisec- 
tion-points of  the  sides  are  concyclic. 

Ex.  616.  If  an  inscribed  equiangular  polygon  have  an 
odd  number  of  sides,  it  is  equilateral. 

Ex.  617.  If  a  circumscribed  equilateral  polygon  have 
an  odd  number  of  sides,  it  is  equiangular. 

Ex.  618.  If  one  of  two  equal  chords  of  a  O  bisects 
he  other,  then  each  bisects  the  other. 

Ex.  619.  The  tri-rect angular  a  is  its  own  polar. 

Ex.  620.  All  =  as  on  the  same  side  of  the  same  base 
have  their  two  sides  bisected  by  the  same  straightest. 

Ex.  621.  If  the  base  of  a  a  be  given,  and  the  vertex 
variable,  the  straightests  through  the  bisection-points 
of  the  two  sides  always  pass  through  two  fixed  points. 

Ex.  622.  If  A  and  A'  be  opposites,  then  as  ABC, 
A'BC  are  called  colunar.  A  pole  of  the  straightest 
bisecting  A B  and  AC  is  also  pole  of  the  circum-O  of  the 
colunar  a  A'BC. 

Ex.  623.  Given  b  and  a  +  r— /?  to  construct  q-pole  and 
radius  of  circum-O- 

Ex.  624.  If  a+  /?  =  r,  the  q-pole  of  circum-o  is  bisection- 
point  of  c. 


246  RATIONAL   GEOMETRY. 

Ex.  625.  Two  as  with  one  ^  the  same  and  the  opposite 
escribed   Os=,have  equal  perimeters. 

Ex.  626  The  tangent  at  A  to  the  circum-O  of  a  ABC 
makes  with  AB  and  AC^s  whose  difference  =/?  —  r- 

Ex.  627.  The  q-pole  of  the  circum-o  of  a  a  coincides 
with  that  of  the  in-O  of  the  polar  *a  ;  and  the  spherical 
radii  of  the  2  O  s  are  complementary. 

Ex.  628.  From  each  /  of  a  a  a  1  is  drawn  to  the 
straightest  through  the  bisection-points  of  the  adjacent 
sides.     Prove  these   ±s  copunctal. 

Ex.  629.  Through  each  ^  of  a  a  a  straightest  is  drawn 
to  make  the  same  ^  with  one  side  as  the  -L  on  the  base 
makes  with  the  other  side.     Prove  these  copunctal. 

Ex.  630.  Two  birect angular  as  are  =  if  the  oblique  ^s 
are  = ,  or  if  the  sides  not  quadrants  are  = . 

Ex.  631.  In  a,  if  c  is  fixed  and  a+/9=w,  then  C  is  on 
a  fixed  straightest. 

Ex.  632.  (Joachimsthal.)  If  two  diagonals  of  a  com- 
plete spherical  quadrilateral  are  quadrants,  so  is  the  third. 

Ex.  633.  (1)  A  quad'  whose  diagonals  bisect  each 
other  (a  cenquad)  has  its  opposite  sides  =  ;  (2)  and  in- 
versely. 

(3)  Also   its   opposite  ^s  = ;    (4)  and  inversely. 

(5)  Every  straightest  through  this  bisection-point 
(spherical  center)  cuts  the  quad'  into  = halves. 

(6)  Its  opposite  sides  make  =  alternate  ^s  with  a 
diagonal. 

(7)  Inversely,  a  quad'  with  a  diagonal  making  with 
each  side  a  }£  =  to  its  alternate  is  a  cenquad.  (8)  So  is 
a  quad  with  a  pair  of  opposite  sides  —  and  making  = 
alternate  ^  s  with  a  diagonal. 

(9)  Also  a  quad'  with  a  pair  of  opposite  sides  — ,  and  a 
diagonal  making  =  alternate  ^  s  with  the  other  sides 
and  opposite  ^  s  not  supplemental. 

(10)  From  the  spherical  center  Is  on  a  pair  of  opposite 
sides  are  =. 


PURE  SPHERICS.  247 

(11)  If  two  consecutive  ^  s  of  a  cenquad  are  =,  it  has  a 
circum-o. 

(12)  If  two  consecutive  sides  of  a  cenquad  are=,it 
has  an  in-O. 

(13)  The  polar  of  a  cenquad  is  a  concentric  cenquad. 

(14)  A  pair  of  opposite  sides  of  a  cenquad  intersect  on 
the  polar  of  its  spherical  center. 

(15)  Any  two  consecutive  vertices  of  a  cenquad  and  the 
opposites  of  the  other  two  are  coney clic. 

(16)  If  ABCD  be  a  cenquad,  then  A,  B,  C,  D'  and  A\ 
B't  C,  D  are  on  =  Os  with  q-poles  opposites. 

Ex.  634.  The  sides  of  a  a  intersect  the  corresponding 
sides  of  its  polar  on  the  polar  of  their  orthocenter. 

Ex.  635.  The  sect  which  a  ^  intercepts  on  the  polar  of 
its  vertex  equals  a  sect  between  poles  of  its  sides. 

Ex.  636.  If  a  spherical  quad'  is  inscribed,  and  another 
circumscribed  touching  at  the  vertices  of  the  first,  the 
crosses  of  the  opposite  sides  of  these  quad's  are  on  a 
straightest . 

Ex.  637.  The  crosses  of  the  sides  of  an  inscribed  a 
with  the  tangents  at  the  opposite  vertices  are  on  a 
straightest. 


CHAPTER  XVI. 

ANGLOIDS  OR  POLYHEDRAL  ANGLES 

570.  Theorem.  The  area  of  a  spherical  angle,  L, 
is  2r2u. 

Proof.  For  we  have  the  proportion,  area  of  £: 
area  of  J  sphere  =  size  of  £ :  size  of  r't  ^=size  of  ^ 
at  center  :  size  of  r't  £ ;   that  is, 

L  :  r27i=u  :  \iz. 
.'.  L  =  2rht. 

571.  Corollary  to  570  and  566.  The  area  of  a 
spherical  triangle  is  the  size  of  its  spherical  excess 
multiplied  by  its  squared  radius. 

If  e'  is  the  u  of  e, 


572.  Corollary  to  571.  To  find  the  area  of  a 
spherical  polygon,  multiply  its  spherical  excess  in 
radians  by  the  squared  radius. 

573.  Definition.  Three  or  more  rays,  a,  b,  c, 
from  the  same  point,  V,  taken  in  a  certain  order 
and  such  that  no  three  consecutive  are  coplanar, 
determine  a  figure  called  a  polyhedral  angle  or  an 
angloid. 

The  common  point  V  is  the  vertex,  the  rays  a,  b, 
c,  .  .  .  are  edges   the  angles   ^ab    tbc,  .  .  .  are  faces, 

248 


ANGLOIDS  OR  POLYHEDRAL   ANGLES.  249 

and  the  pairs  of  consecutive  faces  are  the  dihedrals 
of  the  angloid. 

According  to  the  number  of  the  rays,  3,  4,  5,  .  .  . 
the  angloid  is  called  trihedral,  tetrahedral,  penta- 
hedral, .  .  .  ,  and  in  general  polyhedral. 

574.  If  a  unit  sphere  be  taken  with  the  vertex  of 
the  angloid  as  center,  this  determines  a  spherical 
polygon  whose  angles  are  of  the  same  size  as  the 
inclinations  of  the  angloid's  dihedrals,  while  the 
length  of  each  side  of  the  polygon  is  the  size  of  the 
corresponding  face-angle  of  the  angloid. 

Hence  from  any  property  of  spherical  polygons 
we  may  infer  an  analogous  property  of  angloids. 

For  example,  the  following  properties  of  trihe- 
drals  have  been  proved  in  our  treatment  of  spheri- 
cal triangles: 

I.  Trihedrals  are  either  congruent  or  symmetrical 
which  have  the  following  parts  congruent : 

(1)  Two  face-angles  and  the  included  dihedral. 

(2)  Two  dihedrals  and  the  included  face-angle. 

(3)  Three  face-angles. 

(4)  Three  dihedrals. 

(5)  Two  pairs  of  dihedrals  and  the  face-angles 
opposite  one  pair  equal,  opposite  the  other  pair  not 
supplemental. 

(6)  Two  pairs  of  face-angles  and  the  dihedrals 
opposite  one  pair  equal,  opposite  the  other  pair  not 
supplemental. 

II.  As  one  of  the  face-angles  of  a  trihedral  is 
greater  than  or  equal  to  a  second,  the  dihedral  oppo- 
site the  first  is  greater  than  or  equal  to  that  opposite 
the  second,  and  inversely. 


250  RATIONAL    GEOMETRY. 

III.  Symmetrical  trihedrals  are  equivalent  or 
equivalent  by  completion. 

IV.  Any  two  face-angles  of  a  trihedral  are  together 
greater  than  the  third. 

V.  In  two  trihedrals  having  two  face-angles  respec- 
tively congruent,  if  the  third  is  greater  in  the  first, 
so  is  the  opposite  dihedral,  and  inversely. 

VI.  In  any  trihedral  the  sum  of  the  three  face- 
angles  is  less  than  four  right  angles. 

VII.  In  any  trihedral,  the  sum  of  the  three  dihe- 
drals is  greater  than  two  and  less  than  six  right 
angles. 

In  the  same  way,  defin'ng  a  polyhedral  as  convex 
when  any  polygon  formed  by  a  plane  cutting  every 
face  is  convex,  we  have : 

VIII.  In  any  convex  polyhedral  any  face-angle  is 
less  than  the  sum  of  all  the  other  face-angles. 

Proof.  Divide  into  trihedrals  and  apply  IV  re- 
peatedly. 

IX.  In  any  convex  polyhedral  the  sum  of  the 
face-angles  is  less  than  four  right  angles. 

X.  The  three  planes  which  bisect  the  dihedrals  of 
a  trihedral  are  costraight. 

XI.  The  three  planes  through  the  edges  and 
the  bisectors  of  the  opposite  face-angles  of  a  tri- 
hedral are  costraight. 

XII.  The  three  planes  through  the  bisectors  of 
the  face  angles  of  a  trihedral,  and  perpendicular 
to  these  faces,  respectively,  are  costraight. 

XIII.  The  three  planes  through  the  edges  of  a 
trihedral,  and  perpendicular  to  the  opposite  faces, 
respectively,  are  costraight. 


ANGLOIDS  OR  POLYHEDRAL  ANGLES,  251 

XIV.  If  two  face-angles  of  a  trihedral  are  right, 
the  dihedrals  opposite  are  right. 

Ex.  638.  The  face  angles  of  any  trihedral  are  propor- 
tional to  the  sides  of  its  a  on  any  sphere. 

Ex.  639.  The  area  of  a  a  is  to  that  of  the  sphere  as 
its  spherical  excess  is  to  8  r't  ifs  {e'-.^n). 

Ex.  640.  Find  the  angles  and  sides  of  an  equilateral  a 
whose  area  is  \  the  sphere. 

Ex.  641.  The  angle-sum  in  a  r't  a  is   <  4  r't  ^s. 

Ex.  642.  If  one  of  the  sects  which  join  the  bisection- 
points  of  the  sides  of  a  a  be  a  quadrant,  the  other  two 
are  quadrants. 

Ex.  643.  Cut  a  tetrahedral  by  a  plane  so  that  the  sec- 
tion is  a  ||gm. 

Ex.  644.  To  cut  by  a  plane  a  trirectangular  trihedral 
so  that  the  section  may  equal  any  given  a. 

Ex.  645.  The  base  AC  and  the  area  of  a  a  being  given, 
the  vertex  B  is  concyclic  with  A'  and  C . 

Ex.  646.  Given  a  trihedral;  to  each  face  from  the 
vertex  erect  a  perpendicular  ray  on  the  same  side  as  the 
third  edge;  the  trihedral  they  form  is  called  the  polar 
of  the  given  one. 

If  one  trihedral  is  the  polar  of  a  second,  then  the  second 
is  also  the  polar  of  the  first. 

Ex.  647.  If  two  trihedrals  are  polars,  the  face  angles  of 
the  one  are  supplemental  to  the  inclinations  of  the  corre- 
sponding dihedrals  of  the  other. 

Ex.  648.  If  two  angles  of  a  a  be  r't,  its  area  varies  as 
the  third  ^  . 

Ex.  649.  If  1',  one  minute,  is  one  sixtieth  of  a  degree, 
and  1",  one  second,  is  one  sixtieth  of  a  minute,  find  the 
area  of  a  a  from  the  radius  r,  and  the  angles  a  =20°  9' 
30",  /?=55°  53'  32",  r=n4°  20'  14".         Ans.  o.i8i3r2. 

Ex.  650.  All  trihedrals  having  two  edges  common,  and, 
on  the  same  side  of  these,  their  third  edges  prolongations 
of  elements  of  a  right  cone  containing  the  two  common 
edges,  are  equivalent. 


252  RATIONAL   GEOMETRY. 

Ex.  651.  Equivalent  as  on  the  same  side  of  the  same 
base  are  between  copolar  =  Os. 

Ex.  652.  Find  the  spherical  excess  of  a  a  in  degrees 
from  its  area  and  the  radius. 

Ex.  653.  If  any  angloid  whose  size  is  1,  that  is,  any 
angloid  which  determines  on  the  unit  sphere  a  spherical 
polygon  whose  area  is  1,  be  called  a  steradian,  and  all 
the  angloids  about  a  point  be  together  called  a  steregon, 
then  a  steregon  contains  47:  steradians. 


APPENDIX  I. 

THE  PROOFS  OF  THE  TWO  BETWEENNESS  THEOREMS 
16  AND  17,  TAKEN  FOR  GRANTED  IN  THE  TEXT.* 

575.  Theorem  I.  If  B  is  between  A  and  C,  and  C 
is  between  A  and  D,  then  C  is  between  B  and  D. 

Proof.  Let  A,  B,  C,  D  be  on  a.  Through  C  take 
a  straight  c  other  than  a.  On  c  take  a  point  E 
other  than  C.  On  the  straight 
BE  between  B  and  E  take  F. 
Thus  between  B  and  F  is  no 
point  of  c.  Now  between  A 
and  F  there  can  be  no  point 
of  cy  else  c  would  (by  II  4) 
have  a  point  between  A  and 
B,  since,  by  the  construction  of  F,  c  cannot  have  a 
point  between  B  and  F.  Thus  C  would  be  between 
A  and  B,  contrary  to  our  hypothesis  that  B  is 
between  A  and  C. 

Thus  since  c  cannot  have  a  point  between  A  and 
F,  it  must  (by  II  4)  have  a  point  between  F  and  D. 
Thus  we  have  the  three  non-co-straight  points  F, 
Bt  D,  and  c  with  a  point  between  F  and  D,  and,  by 
construction,  none  between  F  and  B.  Therefore  it 
must  (by  II  4)  have  a  point  between  B  and  D.  So 
C  is  between  5  and  D. 


*  These  proofs  are  due  to  my  pupil,  R.  L.  Moore,  to  whom  I  have 
been  exceptionally  indebted  throughout  the  making  of  this  book. 

253 


254  RATIONAL   GEOMETRY. 

576.  Theorem  II.  If  B  is  between  A  and  C,  and  C 
is  between  A  and  D,  then  B  is  between  A  and  D. 

Proof.  Let  A,  B,  C,  D  be  on  a.  Through  B  take 
a  straight  b  other  than  a.  On  b 
take  a  point  E  other  than  B.  On 
the  straight  CE  between  C  and 
E  take  F.  Thus  between  C  and 
F  is  no  point  of  b.  Then  since 
by  hypothesis  B  is  between  A 
and  C,  therefore  b  must  (by  II  4) 
Fig.  229.  have  a  point  between  A  and  F. 

Thus  we  have  the  three  non-co-straight  points  A, 
F,  A  and  6  with  a  point  between  A  and  F.  There- 
fore b  must  have  (by  II  4)  a  point  between  A  and  D, 
or  between  F  and  D.  But  it  cannot  have  a  point 
between  F  and  D,  for  then  it  must  (by  II  4)  have  a 
point  either  between  F  and  C,  contrary  to  our  con- 
struction, or  else  between  C  and  D,  contrary  to 
Theorem  I,  by  which  C  is  between  B  and  D.  There- 
fore it  has  a  point  between  A  and  D.  So  B  is  be- 
tween A  and  D. 

577.  Theorem.  Ill  Any  four  points  of  a  straight 
can  always  be  so  lettered,  A  BCD,  that  B  is  between  A 
and  C  and  also  between  A  and  D,  and  furthermore  C 
is  between  A  and  D  and  also  between  B  and  D. 

Proof.  We  may  (by  II  3)  letter  three  of  our 
points  B,  C,  D,  with  C  between  B  and  D.  Now  as 
regards  B  and  D,  and  our  fourth  point  A,  either  A 
is  between  B  and  D,  or  B  is  between  A  and  D,or  D 
is  between  A  and  B. 

If  B  is  between  A  and  D,  we  have  fulfilled  the 
hypothesis  of  Theorems  I  and  IL 


APPENDIX  I.  255 

If  D  is  between  A  and  B,  then  interchanging  the 
lettering  for  B  and  D,  that  is  calling  B,  D  and  A  B, 
we  have  the  hypothesis  of  Theorems  I  and  II.  There 
only  remains  to  consider  the  case  where  A  is  be- 
tween B  and  D. 

If  now  C  is  between  D  and  A ,  we  have  fulfilled  the 
hypothesis  of  Theorems  I  and  II,  by  calling  D,  A, 
and  C,  B,  and  A,  C,  and  5,  D. 

If,  however,  A  were  between  C  and  £>  we  would 
have  fulfilled  the  hypothesis  of  Theorems  I  and  II 
by  writing  for  A}  B,  for  D,  A,  and  for  B,  D. 

We  have  left  only  one  sub-case  to  consider,  that 
where  D  is  between  A  and  C. 
This  sub-case  is  impossible. 
Suppose  A  BCD  on  a. 
Through  Ctake  a  straight  c 
other  than  a.  On  c  take  a 
point  £  other  than  C.  On 
the  straight  DE  between  D 
and  E  take  F.  Thus  between 
D  and  F  is  no  point  of  c.  FlG-  23°- 

Then  since  by  hypothesis  C  is  between  B  and  Z>, 
therefore  c  must  (by  II  4)  have  a  point  between  5 
and  F.  Therefore  we  have  the  three  non-costraight 
points  B,  F}  A}  and  c  with  a  point  between  B  and  F. 
Therefore  c  has  (by  II  4)  either  a  point  between  B 
and  A,  or  a  point  between  F  and  A.  But  it  cannot 
have  a  point  between  F  and  A,  else  it  would  (by  II 
4)  either  have  a  point  between  F  and  D,  contrary  to 
our  construction,  or  else  between  D  and  A,  giving 
C  between  D  and  A,  contrary  to  our  hypothesis  D 
between  A  and  C, 


256  RATIONAL   GEOMETRY. 

So  C  would  be  between  B  and  A,  but  this  with  D 
between  A  and  C  gives  (by  Theorem  I)  D  between 
A  and  B,  contrary  to  our  hypothesis  A  between  B 
and  D. 

Thus  there  is  always  such  a  lettering  that  B  is 
between  A  and  C,  and  C  between  A  and  D,  whence 
(by  Theorem  I)  C  is  between  B  and  D,  and  (by 
Theorem  II)  B  is  between  A  and  D. 

578.  Theorem.  ^4r#  A,  B,  C,  D  points  of  a  straight, 
such  that  C  lies  between  A  and  D  and  B  between  A  and 
C,  then  lies  also  B  between  A  and  D,  but  not  between 
C  and  D. 


Fig.   231. 

Proof.  The  points  A  BCD,  in  accordance  with 
577,  have  an  order  in  which  two  are  each  between 
the  remaining  pair  and  of  this  remaining  pair  neither 
is  between  two  others.  But  here  by  hypothesis  C 
and  B  are  between  others.  So  we  reach  the  follow- 
ing arrangements  ACBD,  BBC  A,  ABCD,  DCBA. 
Of  these  arrangements,  however,  the  first  two  do 
not  satisfy  the  hypothesis.  For  in  both  arrang- 
ments  C  lies  between  A  and  B,  which  (by  II  3)  con- 
tradicts the  hypothesis   "  B  between  A  and  67 ' 

In  the  third  and  fourth  arrangement  appears,  by 
577,  that  C  lies  between  B  and  D,  therefore,  by  II  3, 
B  cannot  lie  between  C  and  D. 

579.  Theorem.  Between  any  two  points  of  a  straight 
there  are  always  indefinitely  many  points. 

Proof.  By  II  2,  there  is  between  A  and  B  at 
least  one  point  C\  likewise  there  is  between  A  and 


APPENDIX  I.  257 

C  at  least  one  point  O .  Further,  there  is  within 
AC  at  least  one  point  C",  which  likewise  is  within 
A  B  but  not  within  OB ;  therefore  since  C  lies  within 
OB,  C  cannot  be  identical  with  C.     In  this  way 


liio.  232. 

we  get  ever  new  points  of  AB  without  ever  coming 
to  an  end. 

580.  Theorem.  If  ABCD  is  an  arrangement  of 
four  points  corresponding  to  577,  then  there  is 
besides  this  arrangement  only  still  the  inverse 
which  fulfills  577.  [The  proof  is  essentially  already 
given  in  proving  578.] 

581.  Theorem.  If  any  finite  number  of  points  of 
a  straight  are  given,  then  they  can  always  be  ar- 
ranged in  a  succession  A,  B,  C,  D,  E,  .  .  .  ,  K,  such 
that  B.  lies  between  A  on  the  one  hand  and  C,  D, 
E,  .  .  .  ,  K  on  the  other,  further  C  between  A}  B  on 
one  side  and  D,  E,  .  .  .  ,  K  on  the  other,  then  D 
between  A,  B,  C  on  the  one  side  and  E,  .  .  .  ,  K  on 
the  other,  and  so  on. 

Besides  this  distribution  there  is  only  one  other, 
the  reversed  arrangement,  which  is  of  the  same 
character. 

[This  theorem  is  a  generalization  of  577.] 


Fig. 


'■33- 


Proof.     Our  theorem  holds  for  four  points  by  577 
and  580, 


258  RATIONAL   GEOMETRY. 

We  may  show  that  the  theorem  remains  valid  for 
n  + 1  points  if  it  holds  for  n  points. 

Let  A1A2A3 .  .  .  An  be  the  desired  arrangement 
for  n  points.  If  further  we  take  an  additional  point 
then  there  are  forthwith  three  cases  possible : 

(1)  At  lies  between  X  and  An\ 

(2)  An  lies  between  X  and  A1\ 

(3)  X  lies  between  At  and  An. 

In  the  third  case  we  prove  further,  that  there  is 
one  and  only  one  number  m,  such  that  X  lies  between 
A  m  and  Am+l. 

Finally  we  show  that  the  following  arrangements 
in  the  three  cases  have  the  desired  properties: 

(1)  XA,A2A3.  ..An; 

(2)  A^A^^AnX; 

(3)  ^-1-^2^3  •  •  •  AmXAm+l  .  .  .  An\ 

and  that  they  with  their  inversions  are  the  only  ones 
which  possess  those  properties. 


APPENDIX  II. 


THE     COMPASSES 


582.  Euclid's  third  postulate  is:  About  any  cen- 
ter with  any  radius  one  and  only  one  circle  may  be 
taken.  This  has  been  understood  in  ordinary  geom- 
etries as  authorizing  the  use  of  a  physical  instrument, 
the  compasses,  for  drawing  a  circle  with  any  center 
and  any  radius. 

But  this  is  only  made  fruitful,  beyond  the  sect- 
carrier,  in  problem  solving,  by  two  new  assumptions : 

Assumptions  of  the  Compasses. 

Assumption  VI  1 .  If  a  straight  have  a  point  within 
a  circle,  it  has  two  points  on  the  circle. 

Assumption  VI  2.     If  a  circle  have  a  point  within 
and  a  point  without  another 
circle,  it  has  two  points  on  this 
other.  1 

583.  Problem.  From  a  given  \ 
point  without  the  circle  to  \ 
draw  a  tangent  to  the  circle. 

Construction.  Join  the 
given  point  A  with  the  cen- 
ter C,  meeting  the  circle  in 
B.     Erect  BD±_   to   CB,  and  (by  VI   1)  cutting  in 

259 


\ 


260  RATIONAL   GEOMETRY. 

D  the  OC(CA).  Join  DC,  meeting  OC(CB)  in  F. 
Then  AF  is  tangent  to  OC(CB). 

Proof.  Radius  CA,  _L  to  chord  HD,  bisects  arc 
HD\  .'.  if  we  rotate  the  figure  until  H  comes  upon 
the  trace  of  A,  then  A  is  on  the  trace  of  D;  .'.  tan- 
gent HB  en  trace  of  AF. 

Determination.  Always  two  and  only  two  tan- 
gents. 

584.  Problem.  To  construct  a  triangle  of  which 
the  sides  shall  be  equal  to  three  given'  sects,  given 
that  any  two  whatever  of  these  sects  are  together 
greater  than  the  third. 


Fig.  235. 

Given  the  three  sects  a,  b,  c,  any  two  whatever 
together  greater  than  the  third. 

Construction.  On  a  straight  OF  from  0  take 
OG  =  b.  Take  O0(a),  and  OG(c).  Since  a  +  c>b, 
these  (by  VI  2)  intersect,  say  at  K. 

a  OGK  is  the  triangle  required. 

585.  Problem.  To  construct  a  triangle,  given  two 
sides  and  the  angle  opposite  one  of  them. 

Given  a,  c,  and  C. 


APPENDIX  II. 


26 


Case  i.     If  a<c. 

On  one  ray  of  ^-C  take  CB  =  a. 
This  (by  VI  1)  has  two  points  A' , 
Ax  on  the  straight  of  the  other  ray 
of  ^  C.  The  point  C  is  between  Af 
and  Av  .*.  if  ^C  r't,  we  have  two 
congruent  triangles  (Fig  236) ;  if 
oblique,  only  one  triangle  (Fig.  237) 


Take  OB(c). 

B 


Fig.  236. 


Case  2.    Iia=c.    [^fC  acute.] 
Then  C  coincides  with  A'  or  Alt 
and  we  have  only  one  triangle. 
Fig.  237. 

Case  3.    If  a>c.    [^fC  acute. J 

I.  If  c  =  p,  the  perpendicular  from  B  on  CA,  there 
is  only  one  triangle. 

II.  If  c>p,  then  At  and  Af  are  on  the  same  side 
of  C  and  there  are  two  different  B 
triangles  which  fulfill  the  condi- 
>ions,  namely,  A'BC  and  AtBC 
(Fig.  238). 

This  is  called  the  ambiguous 
case. 

III.  If  c<p,  no  triangle. 


Fig.  238. 


APPENDIX  III. 

THE  SOLUTION  OF  PROBLEMS. 

586.  A  problem  in  geometry  is  a  proposition  ask- 
ing for  the  graphic  construction  of  a  figure  which 
shall  satisfy  or  fulfill  certain  given  conditions  or 
requirements.  It  has  been  customary  to  use  the 
ruler  and.  compasses ;  that  is,  to  allow  our  assump- 
tions I-V  and  also  VI  (Appendix  II),  but  no  others. 
Of  these,  assumptions  V-VI  have  usually  been 
superfluous  and  unnecessary,  the  problems  treated 
not  requiring  the  compasses,  but  only  ruler  and  sect- 
carrier. 

587.  When  we  know  how  to  solve  a  problem,  the 
treatment  consists  of 

(1)  Construction:  Indicating  how  the  ruler  and 
sect-carrier  or  ruler  and  compasses  are  to  be  used 
in  effecting  what  is  required. 

(2)  Proof:  Showing  that  the  construction  gives  a 
figure  fulfilling  all  the  requirements. 

(3)  Determination:  Considering  the  possibility  of 
the  solution,  and  fixing  whether  there  is  only  a 
single  solution  or  suitable  result  of  the  indicated 
procedure,  or  more  than  one,  and  discussing  the 
limitations  which  sometimes  exist,  within  which 
alone  the  solution  is  possible. 

262 


APPENDIX  III  263 

588.  The  first  step  toward  finding  a  desired  solu- 
tion is  usually  what  is  called  Geometrical  Analysis. 

This  consists  in  supposing  drawn  a  figure  like  the 
one  desired,  also  containing  the  things  given,  and 
then  analyzing  the  relations  of  the  given  things 
among  themselves  and  to  the  things  or  figure  sought, 
or  the  elements  necessary  for  attaining  such  figure. 

589.  Methods  of  procedure  in  problem-solving. 

I.  Successive  Substitutions. — We  may  substitute 
for  the  required  construction  another  from  which  it 
would  follow,  and  for  this  another,  perhaps  simpler, 
until  one  is  reached  which  we  know  how  to  accom- 
plish. 

Just  so,  in  attempting  to  find  a  demonstration  for 
a  new  theorem,  we  may  freely  deduce  from  the  de- 
sired proposition  by  use  of  invertible  theorems,  and 
if  thus  we  reach  a  known  proposition,  the  inversion 
of  the  process  will  give  the  demonstration  sought. 


Fig.  239. 

I  Example  1 .  Theorem.     If  from  any  point  P  on 
the  circumcircle  of  the  triangle  ABC  be  drawn  PX, 


264 


RATIONAL    GEOMETRY. 


PY,  PZ  perpendicular  to  the  sides,  the  points  X, 
y,  Z  will  be  eostraight,  the  Simson's  st'  of  a  for  P. 
Analysis.  We  will  have  proven  X,  y,  Z  eostraight 
if  on  joining  XY,  XZ  we  show  4-PXY  supplement- 
ary to  ^  PXZ.  But  again  this  is  proven  if  we  show 
■4.PXZ  supplementary  to  ^ABP  and  i-PXY  = 
fABP. 

But  tPXZ  is  supplement  of  ^ABP  since  P,  X, 
Z,  B  are  coney clic.  And  since  P,  Y,  C,  X  are  con- 
cyclic,  £PXY  is  supplement  of  ^-PCY}  as  is  also 
£ABP,  since  P,  C,  A,  B  are  concyclic. 

I  Example  2.  Problem.  Construct  a  a,  given  an 
angle,  the  side  opposite  and  two  sects  proportional 
to  the  other  two  sides,     [a  from  a,  a,  b/c] 

Analysis.  By  a  and  a  is  (by  165)  the  circumcircle 
given.  Bisect  4.BAC  by  AD 
and  prolong  AD  to  meet  the 
circle  again  (by  138)  in  E. 
Then  (by  242)  CD/BD  =  b/c. 
So  (by  241)  the  point  D  is 
known.  Moreover,  since  arc 
BE  =  arc  CEy  the  point  E  is 
(by  225)  known.  Therefore, 
taking  BC  =  a,  the  points  D 
and  E  can  be  constructed,  and 
thus  A  by  the  prolongation  of  ED  to  the  circle. 

Analogous  Problems:  Ex.  1.  Given  a,  P,  b/c. 
Ex.  2.  Given  a,  R,  b/c.     Ex.  3.  Given  a,  BD,  CD. 

II.  Data. — The  explicit  giving  of  certain  things 
may  involve  the  implicit  giving  of  others  more  im- 
mediately available. 


Fig.  240. 


APPENDIX  III.  265 

Such  an  implicitly  given  thing  has  been  called  a 
datum. 

For  example,  if  a  straight  and  a  point  without  it 
be  given,  the  perpendicular  from  the  point  to  the 
straight  is  a  datum. 

If  an  angle  and  a  point  within  it  be  given,  then 
the  sect  from  the  point  to  the  vertex,  the  sect  from 
the  point  to  one  side  drawn  parallel  to  the  other 
side,  and  the  sect  this  cuts  off  from  the  side  are  data. 

With  an  angle  a  are  given  the  constructible  parts 
\a,  \a,  etc.,  but  not  \a,  \a,  etc.;  also  the  supple- 
ment and  complement. 

If  the  sum  and  difference  of  two  magnitudes  are 
given  so  are  the  magnitudes. 

If  in  a  triangle  of  the  three  things,  a  side,  the  oppo- 
site angle,  the  circumradius,  two  are  given,  so  is  the 
third. 

So  also  with  base,  altitude,  area. 

II  Example  1.  To  construct  a  triangle  from  one 
side,  the  opposite  angle,  and  the  difference  of  the 
other  two  angles,     (a  from  a,  a,  ft—  y.) 

Analysis.  Since  a  is  known,  so  is  also  /?+  y  as  its 
supplement.  .*.  /?  and  y  are  known,  and  we  have  a 
side  and  the  two  adjoining  angles. 

III.  Translation. — Again  new  auxiliary  parts  may 
advantageously  be  introduced.  Certain  procedures 
are  found  particularly  fertile. 

In  any  triangle  ABC,  transporting  AC  parallel  to 
itself  into  BD,  and  extending  B A  equal  to  itself  to 
E,  we  have  that  the  sides  of  EDC  are  double  the 
medians  of  ABC  and  parallel  to  them. 


266  RATIONAL   GEOMETRY. 

The  sides  of  ABC  are  two-thirds  the  medians  of 
EDC,  and  A  is  its  centroid.  Two  of  the  altitudes 
of  the  triangles  A  ED,  A  EC,  ADC  are  equal  to  two 
altitudes  of  ABC  and  the  content  of  EDC  is  triple 
that  of  ABC. 

If  we  have  given  such  elements  of  ABC  as  render 
possible,  through  these  properties,  the  determina- 


Fig.  241. 

tion  of  one  of  the  triangles  EDC,  A  ED,  ADC,  A  EC, 
then  the  triangle  ABC  will  always  be  constructible. 
Ill  Example  1.  Problem.  Construct  a  triangle, 
given  one  median  mx,  the  angle  between  the  others, 
m2  and  m3,  and  two  sects  proportional  to  them : 

[a  from  mlt  ^of  m2  and  m3,  m2/m3]. 

Analysis.  In  the  a  EDC  we  know  DC  =  2FC  = 
2Wt ;  also  %E  and  DE/EC.  Therefore  the  problem 
is  reduced  to  I  Example  2. 

IV.  Symmetry. — Add  to  the  tentatively  con- 
structed figure  the  figure  symmetrical  to  it  or  to  a 
part  of  it,  with  respect  to  a  chosen  straight  as  axis. 
Particularly  adapted  for  axis  is  an  angle-bisector 
or  a  perpendicular.  Especially  does  this  show 
differences  of  sects  or  angles. 


APPENDIX  III. 


267 


IV  Example 


Problem.  To  construct  a  tri- 
angle, given  two  sides  and  the 
difference  of  the  opposite  an- 
gles: 

[a  from  a,  b,  a—fl. 


E  D 

Fig.  242. 


Analysis.         Take   CE-vCA 
with  respect  to  perpendicular 
CD.      Then  $AEC  =  a.     But  *AEC  =^B+^  BCE 
=(3+d.     .-.*•*«-£ 

A  BCE  is  therefore  constructive  (two  sides  and 
included  £). 

The  point  D  is  the  foot  of  the  perpendicular  from 
C  on  BE.     A  +  E  (axis  CD). 

IV  Example  2.     Given  the  two  straights  s  and  I 
and  a  point  A  on  s.     Determine 
the  point  X  on  s,  such  that  AX  = 
perpendicular  from  X  on  /. 

Analysis.  With  respect  to  the 
bisector  of  ifAXB  as  axis,  the 
figure  symmetrical  to  XBC  is  XACt 
and  $ACB  is  bisected  by  XC.  Therefore,  erecting 
a  perpendicular  at  the  given  point  A,  the  bisector 
of  the  angle  ACB  made  with  /  meets  5  in  the  required 
point  X. 

V.  Similarity. — When  the  figure  to  be  constructed 
is  determinable  by  such  conditions  that,  omitting 
one,  the  remaining  determine  a  system  of  similar 
figures,  then  first  determine  this  system  of  figures 
similar  to  the  one  sought,  and  secondly,  by  taking 
account  of  the  suppressed  condition,  fix  the  one 
wMch  solves  the  problem. 


Fig.  243- 


268  RATIONAL   GEOMETRY. 

There  are  two  sorts  of  cases  according  as  the  sup- 
pressed condition  determines  (I)  the  size  of  the 
figure  sought;  (II)  its  position  or  place.  In  the 
first  case  the  data  are  angles  and  proportions  (giv- 
ing the  system  of  similar  figures) ,  and  a  sect  (giving 
the  size  of  the  particular  figure) .  In  the  second  case 
are  only  given  angles  and  proportions,  but  also  is 
imposed  a  condition  that  the  figure  demanded  must 
have  a  determinate  position  with  respect  to  some- 
thing given;  for  example,  must  contain  a  given 
point,  must  contain  a  tangent  to  a  given  circle,  etc. 

V  Example  i.  To  make  a  triangle,  given  an 
angle  a,  sects  proportional  to  the  sides  containing 
it  (b:c  =  tn:n),  and  also  its  bisector  tt: 

[A  from  a,  b/c,  tx]. 

On  the  sides  of  a  take  m  and  n.  On  its  bisector 
take  tv  Through  the  foot  of  tx  draw  a  parallel  to 
the  straight  through  the  ends  of  m  and  n.  [Similarly : 
A  from  a,  p,  ma\   a  from  a,  /?,  ha]. 

V  Example  2.  In  a  triangle  ABC  inscribe  a 
parallelogram  of  which  one  angle  shall  coincide  with 
angle  BAC,  and  such  that  its  sides  are  as  m  to  n. 

The  vertices  of  the  parallelogram  sought  must  be 
one  at  A,  one  on  6,  one  on  c,  one  on  a.  Omitting 
this  last  condition,  the  fourth  vertices  of  parallelo- 
grams satisfying  the  other  conditions  are  on  a 
straight  determined  by  A  and  the  parallelogram 
with  sides  m  and  n. 

The  fourth  vertex  sought  is  where  this  straight 
crosses  the  base  6. 


APPENDIX  III.  269 

VI.  Intersection  of  Loci. — All  points  in  a  plane 
which  satisfy  a  single  geometric  condition  make  up 
often  a  single  straight  or  a  single  circle,  in  rare  cases 
more  than  one. 

Neglecting  these  rare  cases,  we  may  call  such 
straight  or  circle  the  locus  (place)  of  the  points  satis- 
fying the  given  condition. 

Where  it  is  required  to  find  points  satisfying  two 
conditions,  if  we  leave  out  one  condition,  we  may 
find  a  locus  of  points  satisfying  the  other  condition. 

Thus,  for  each  condition  we  may  construct  the 
corresponding  locus.  If  these  two  loci  have  points 
in  common,  these  points,  and  these  only,  satisfy  both 
conditions. 

In  a  problem  involving  more  than  two  distinct 
conditions,  two  may  be  selected  which  give  avail- 
able loci,  and  then  the  remaining  used  to  complete 
the  solution.  If  the  circle  occurs  as  locus,  we  may 
assume  the  two  postulates  of  the  compasses  (VI). 

As  preliminary  it  will  be  convenient  to  have  a 
collection  of  simple  loci. 

Loci. 

1 .  The  locus  of  points  which  with  a  given  point  P 
give  the  sect  r  is  QP(r). 

This  is  also  the  locus  of  the  centers  of  circles  with 
radius  r,  which  pass  through  P. 

2.  The  locus  of  points  P  on  one  side  of  a  st'  /  and 
such  that  from  them  the  perpendiculars  on  /  are 
equal  to  h  is  the  parallel  to  /  through  P,  one  such 
point. 


27°  RATIONAL   GEOMETRY. 

This  is  also  the  locus  of  the  centers  of  circles  with 
radius  r  tangent  to  /  on  one  side. 

3.  The  locus  of  the  point  to  which  sects  from  two 
given  points  are  equal  is  the  perpendicular  bisector 
of  the  sect  joining  them. 

This  is  also  the  locus  of  the  centers  of  all  circles 
through  the  two  given  points. 

4.  The  locus  of  the  vertices  of  all  right  angles  on 
the  same  side  of  a  given  sect  as  hypothenuse  is  the 
semicircle  on  the  given  sect  as  diameter. 

5.  The  locus  of  the  vertices  of  all  angles  congruent 
to  p  on  a  given  sect  AC  as  base  is  the  arc  on  AC  hav- 
ing /?  as  inscribed  angle: 

This  is  the  locus  of  the  vertex  of  triangles  on  the 
same  side  of  given  base  b  with  given  opposite  angle  /?. 

6.  The  locus  of  the  point  from  which  perpendic- 
ulars on  the  sides  of  a  given  angle  are  equal  is  the 
angle-bisector. 

This  is  also  the  locus  of  the  centers  of  circles 
touching  both  sides  of  the  angle  from  within. 

7.  The  locus  of  the  bisection-points  of  all  chords 
equal  to  k  in  a  given  circle  is  a  concentric  circle  with 
radius  equal  to  the  perpendicular  from  the  center 
on  k. 

8.  The  locus  of  the  bisection-points  of  all  chords 
of  a  circle  through  a  given  point  P  is  the  circle  on 
the  sect  from  P  to  the  given  center  as  diameter. 

9.  The  locus  of  the  centers  of  circles  touching  a 
given  st'  I  at  the  point  P  is  the  perpendicular  to  / 
through  P. 

10.  The  locus  of  the  centers  of  circles  touching 
the  given  circle  OC(CP)  at  P  is  the  st'  CP. 


APPENDIX  III.  271 

1 1 .  The  locus  of  the  centers  of  circles  with  radius 
rx  touching  a  given  circle  with  radius  r2  from  without 
is  a  concentric  circle  with  radius  rx  +  r2.  From 
within,  radius  r2-rv 

1 2 .  The  locus  of  the  end-points  of  tangents  =  t  to 
the  circle  OC(r)  is  the  concentric  circle  with  radius 
from  C  to  the  end  of  one  of  these  tangents. 

This  circle  is  also  the  locus  of  the  centers  of  cir- 
cles with  radius  t  which  cut  OC(r)  at  right  angles; 
that  is,  so  that  at  every  intersection-point  the  tan- 
gents to  the  two  circles  are  at  right  angles. 

13.  The  locus  of  the  centers  of  circles  of  radius  r 
with  centers  on  the  same  side  of  st'  /  and  cutting 
from  I  a  chord  =  k  is  a  parallel  to  /  through  the  vertex 
of  an  isosceles  triangle  with  base  k  on  /  and  side  r. 

14.  The  centers  of  circles  of  radius  r,  cutting  from 
a  given  circle  OC(r2)  an  arc  with  chord  =k  lie  on 
two  concentric  circles  with  radius  from  C  to  the 
vertex  of  an  isosceles  triangle  with  base  k  a  chord  of 
OC(r2)  and  side  rx. 

15.  The  locus  of  the  centers  of  circles  with  radius 
r,  which  bisect  a  given  circle  OC(r2)  is  a  concentric 
circle  with  radius  from  C  to  the  vertex  of  an  isos- 
celes triangle  with  base  a  diameter  of  OC(r2)  and 
side  rv 

16.  The  locus  of  the  centers  of  circles  with  radius 
rx  which  are  bisected  by  a  given  circle  OC(r2)  is  a 
concentric  circle  having  as  radius  the  perpendicular 
from  C  on  any  chord  =  2^  in  oC(r2). 

17.  The  locus  of  the  vertices  of  triangles  of  equal 
content  on  the  same  side  of  the  same  base  is  the 
parallel  to  the  base  through  the  top  of  its  altitude. 


272 


RATIONAL    GEOMETRY. 


18.  The  locus  of  the  vertex  B  of  all  triangles  on 
the  same  base  b  in  which  a2  +  c2  =  constant  is  a  circle. 

19.  The  locus  of  the  vertex  B  of  all  triangles  on 
the  same  base  b  in  which  a2  —c2  =  constant  is  a  per- 
pendicular to  the  base. 

20.  The  locus  of  the  bisection-point  of  a  sect  with 
end-points  on  two  straights  at  right  angles  is  a  circle 
with  their  intersection  as  center  and  half  the  sect 
as  radius. 

VI  Example    1.     To  construct   a  triangle  from 
an  angle  and  the  altitudes 
on  the  including  sides  (A 
from  a,  h2,  h3). 

Analysis.  B  is  the  inter- 
section of  a  side  of  a  with 
the  parallel  to  the  other  side 
of  a  through  the  end  of  a 
perpendicular  to  this  side  equal  to  h2. 

VI  Example  2.  To  con- 
struct a  triangle  from  one 
side  and  the  altitudes  to 
the  other  two   (a  from  a, 

Analysis.  E  is  the  inter- 
section of  semicircle  on  BC 
with  arc  of  radius  =  h2.  So 
for  F.  Then  A  is  intersec- 
tion of  CE  and  BF. 

VII.  Reckoning.  —  Our 
sect  calculus  may  be  freely  used  for  making  and 
solving  equations  of  the  first  and  second  degree 
containing  expressions  for  sought  sects. 


Fig.  244. 


Fig. 


245' 


APPENDIX  III. 


273 


VII  Example  1.  Construct,  without  using  the 
compasses,  an  isosceles  triangle 
with  the  equal  angles  double  the 
third. 

Construction.      On  one  side  of  a 
D  right  angle,  B,  take  BA  equal  to 
half   the   unit  sect.     BA=%.     On 
the    other    side,  BC  =  {.     Prolong 
the  hypothenuse  AC  to  D,  taking 
Fig.  246.         CD  =  J.     At  D  erect  a  perpendi- 
cular, and  on  it  take  DF  =  \. 

On  CA  take  CE  =  \.  Take 
AA'  =  2AE.  At  E  erect  the 
perpendicular  bisector  EG, 
taking  EG=AF.  Join  AG 
and  A'G.  The  triangle  AA'G 
is  the  one  required. 

Proof.       Take     GH  =  AA'. 
Join  A'H. 
fer-AF-Qa+AP+iP  +  tf 

=  [[i[(5)*+l]]a  +  i]*  FlG'247- 

=  [TV[io  +  2(5)*]]*. 

AE  =  {[(5y -i]. 
.•.^2  =  1V[io  +  2(5)^]+TV[6-2(5^  =  i. 
GH=AA'  =  2AE  =  i[(sy-i]. 

AH  =  i-i[(sy-i]=i[3-(syi 
/.GA:AA'=AA':AH. 
.-.(by  239)  aAGA'~aAA'H. 
;.A'H=AAf  =GH. 


274  RATIONAL    GEOMETRY. 

.\  tAA'G=if.A'AG=tAHA'  = 

=  $  FLGA'  +  4  YLA'G  -  2  *  #GA'. 

VIII.  Partition  of  a  Perigon. — The  four  right 
angles  around  a  point,  taken  together,  may  be  called 
a  perigon.  Since  any  angle  may  be  bisected,  a  peri- 
gon can  be  cut  into  2"  congruent  angles. 

Since  the  supplement  of  the  angle  of  an  equilateral 
triangle  is  one-third  of  four  right  angles,  therefore  a 
perigon  can  be  cut  into  3-2"  congruent  angles. 

The  angles  at  the  base  of  an  isosceles  triangle  with 
the  equal  angles  each  double  the  third  are  each  one- 
fifth  of  four  right  angles.  Therefore  a  perigon  can 
be  cut  into  5  •  2n  congruent  angles. 

The  difference  between  one-third  and  one-fifth  of 
a  perigon  is  two-fifteenths  of  a  perigon.  Hence  a 
perigon  can  be  cut  into  15  -2n  congruent  angles. 

If  a  perigon  be  cut  into  n  congruent  angles,  the 
rays  determine  on  any  circle  about  the  vertex  the 
vertices  of  an  inscribed  regular  polygon  of  n  sides, 
and  the  points  of  tangency  of  a  regular  circum- 
scribed polygon  of  n  sides. 

From' the  time  of  Euclid,  about  300  b.c,  no  ad- 
vance was  made  in  the  inscription  of  regular  poly- 
gons until  Gauss,  in  1796,  found  that  a  regular 
polygon  of  17  sides  was  inscriptible,  and  in  1801 
published  the  following: 

That  the  geometric  division  of  the  circle  into  n 
equal  parts  may  be  possible  it  is  necessary  and  suffi- 
cient that  n  be  2  or  a  higher  power  of  2,  or  else  a 
prime  number  of  the  form  2 2m  + 1 ,  or  a  product  of 
two  or  more  different  prime  numbers  of  that  form, 


APPENDIX  III.  275 

or  else  the  product  of  a  power  of  2  by  one  or  more 
different  prime  numbers  of  that  form.  Below  300, 
the  following  38  are  the  only  possible  values  of  n: 
2,  3,  4,  5,  6,  8,  10,  12,  15,  16,  17,  20,  24,  30,  32,  34, 
40,  48,  51,  60,  64,  68,  80,  85,  96,  102,  120,  128,  136, 
160,  170,  192,  204,  240,  255,  256,  257,  272. 

There  is  only  one  inscriptible  regular  polygon 
known  with  the  number  of  its  sides  prime  and 
greater  than  257.  This  number  is  224 +1  =65,537. 
For  m  =  5,  m=6,  m  =  j,  the  numbers  obtained  are 
not  prime.     Further  no  one  has  gone. 

Ex.  654.  Show  how  to  trisect  the  central  ^,  the  in- 
terior ~4- ,  and  the  ( xterior  ^  of  a  regular  w-gon  where 
n  =2m  or  5.2"*. 

Ex.  655.  Show  how  to  cut  the  %■  o"  an  equilateral 
A  into  n  equal  parts  if  n  is  2'"  or  3.2'"  or  5.2'". 

Loci. 
Ex.  656.  In   a,  given  b  and  p,  find  the  locus  of  G\    H; 

I;  /»;  O. 

Ex.  657.   In    QC(r)  find  the  locus  of  C: 

(1)  Given  r  and  P  (a  point  of  O). 

(2)  Given  r  and  a  tangent. 

(3)  Given  P,  Q  (two  points  of  o). 

(4)  Given  tangent  at  P. 

(5)  Given  2  ||  tangents. 

Ex.  658.  In  a,  given  b  and  /?,  find  locus  of  the  bisection- 
point  of  sect  joining  outer  vertices  of  equilateral  As  on 
a  and  c. 

Ex.  659.  Find  locus  of  point  the  sum  of  the  squares  of 
whose  sects  to  A,  B,  C  =  k. 

Ex.  660.  Given  an  equilateral  a,  find  the  locus  of  the 
point  whose  sect  to  one  vertex  is  the  sum  of  its  sects  to 
the  others. 

Ex.  661.  Find  the  locus  of  the  point  the  sum  of  the 
squares  of  whose  Is  to  the  sides  of  a  r't   ^  =  k2. 


276  RATIONAL   GEOMETRY. 

Ex.  662.  Find  the  locus  of  the  intersection  of  two 
secants  drawn  through  the  ends  of  a  fixed  diameter  in 
a  given  0 ,  one  of  the  secants  1  toa  tangent  at  the  second 
point  where  the  other  cuts   O. 

Ex.  663.  Find  the  locus  of  the  intersection  of  2  st's 
drawn  from  the  acute  ^s  of  a  r't  a,  through  the  points 
where  any  j_  to  hypothenuse  cuts  one  opposite  side  and 
the  production  of  the  other. 

Ex.  664.  Two  given  ©s  intersect.  Find  the  locus  of 
the  bisection-point  of  the  sect  through  one  of  their  points 
of  intersection  with  end -points  one  on  each  circle. 

Ex.  665.  Given  AB  divided  at  C.  Find  locus  of  P, 
if  ^APC^^BPC. 

Ex.  666.  Any  2±  chords  intersect  in  a  given  point 
of  a  given  ©.  Find  the  locus  of  the  bisection-point  of  a 
chord  joining  their  ends. 

Ex.  667.  The  locus  of  a  point,  the  sum  of  the  squares 
of  whose  sects  from  the  vertices  of  a  given  equilateral  a 
equals  twice  the  square  on  one  of  the  sides,  is  the  circum-©. 

Ex.  668.  The  locus  of  the  end  of  a  given  sect  from  the 
point  of  contact  and  on  the  tangent  is  a  concentric  ©. 

Ex.  669.  Find  the  locus  of  the  foot  of  the  _L  from  P  on 
a  st'  through  B. 

Ex.  670.  Find  the  locus  of  the  end  of  sect  from  P  cut 
by  st'  a  into  parts  as  m  to  n. 

Ex.  671.  Find  locus  of  end  of  sect  from  st'  a  cut  by  P 
into  parts  as  m  to  n. 

Ex.  672.  Sects  ||  and  with  ends  in  the  sides  of  if  a  are 
cut  into  parts  as  m  to  n.  Find  the  locus  of  the  cutting 
points. 

Ex.  673.  Find  the  locus  of  a  point  P  if  PA.PB  =m:n. 

Ex.  674.  The  locus  of  the  cross  of  two  tangents  to 
©C(r),  the  st'  of  whose  chord  of  contact  rotates  about  a 
fixed  point  P  is  a  st'  p±CP. 

P  is  called  the  pole  of  p,  and  p  the  polar  of  P,  with 
respect  to  the  given  ©. 

Ex.  675.  If  A  (given)  is  on  the  polar  of  X  (variable), 
find  the  locus  of   X. 


APPENDIX  III.  277 

Ex.  676.  To  find  the  locus  of  a  point  from  which  rays 
through  the  ends  of  a  given  sect  make  a  given   ^ . 

Ex.  677.  Find  the  locus  of  the  vertex  B  of  As,  given 
b  and  a:c  =m:n. 

Ex.  678.  [Circle  of  Apollonius.]  If  a  sect  is  cut  into 
parts  as  m  to  n,  and  the  interior  and  exterior  points  of 
division  are  taken  as  ends  of  a  diameter,  this  0  contains 
the  vertices  of  all  as  on  the  given  sect,  whose  other  two 
sides  are  as  m  to  n. 

Ex.  679.  Find  the  locus  of  those  points  in  a  plane  a 
from  which  rays  to  the  ends  of  a  given  sect  not  on  a  are  _L . 

Ex.  680.  Find  the  locus  of  P  if  PA  =PB  =PC. 

Ex.  681.  If  b'  is  the  projection  of  b  on  a\\b,  and  b' La 
in  a,  find  the  locus  of  the  bisection-point  of  a  given  sect 
AB  if  A  on  a  and  B  on  6. 

Ex.  682.  A  variable  st'  is  ||  to  a  given  plane  and  meets 
two  non-coplanar  st's.  Find  the  locus  of  a  point  which 
cuts  the  intercepted  sect  into  parts  as  m  to  n. 

Ex.  683.  Find  the  locus  of  the  point  from  which  -Ls 
to  three  coplanar  st's  are  = . 

Ex.  684.  Find  the  locus  of  the  point  having  one  or  two 
of  the  following: 

I)  Equal  sects  to  two  given  points; 
(II)  Equal  i.s  to  two  given  intersecting  st's; 

(III)  Equal   _Ls  Jo  two  given  planes. 

Ex.  685.  Find  the  locus  of  the  poles  of  great  circles 
making  a  given  angle  with  a  given  great  circle. 

Ex.  686.  Calling  2  3-s  csmplemental  wh  n  their  sum  is 
a  r't  ^  ,  what  is  the  locus  of  the  intersection  of  rays  from 
A  and  B  making  ~4-  with  AB  the  complement  of  ^  with 
BA? 

Ex.  687.  Calling  a  chord  the  chord  of  contact  of  the 
point  of  interesection  of  tangents  at  its  extremities,  what 
is  the  locus  of  points  whose  chords  of  contact  in  QC(r) 
equal  r? 

Ex.  688.  The  locus  of  vertex  of  a  =s2,  on  given  6, 
is  st'  ||  b  at  altitude  lib,  where  bhb  =2s2. 


278  RATIONAL   GEOMETRY. 

Ex.  689.  The  locus  of  vertex  of  A  on  given  b,  and 
with  a2-c2=s2,  is  st'  1  to  6  at  D,  where  AD2  -CD2  =s\ 

Ex.  690.  Find  locus  of  trisection  points  of  equal  chords. 

Ex.  691.  The  locus  of  a  point  from  which  tangents  to 
two  given  Os  are  =  is  a  st'  _L  to  the  center  sect,  which 
so  divides  it  that  the  difference  of  the  sq's  of  the  seg- 
ments =r2  —  r2.  This  st'  is.  the  radical  axis  of  the  2  0s.  If 
they  intersect  it  contains  their  common  chord. 

Ex.  692.  The  locus  of  P  such  that  PA  :  PB  =  m  :  n  is  o 
on  DjDi  as  diameter,  where  DAB  co-st'  and  DA  :  DB  -tn;n. 

Ex.  693.  Given  b  and  a  —  c,  the  locus  of  foot  of  _L 
from  A  on  tb  is  O  with  bisection-point  of  b  for  center  and 
%{a  —c)  for  radius. 

Ex.  694.  Given  b  and  a  +c,  the  locus  of  foot  of  ±  from 
A  on  bisector  of  external  %■  at  B  is  O  with  bisection- 
point  of  b  for  center  and  J(d  +c)  for  radius 

Ex.  695.  The  locus  of  P  cutting  sects  from  A  to  a  as 
m  to  n  is  a  st'  ||  a. 

Ex.  696.  The  locus  of  P  cutting  a  sect  s  from  A  to  a 
so  that  s-AP=k2  is  a    O. 

Ex.  697.  The  locus  of  P  cutting  a  sect  from  OC(r)  to 
A  as  m  to  n  is  a  O. 

Ex.  698.  If  rectangles  have  one  vertex  at  A  and  the 
adjacent  vertices  on  ©C(r),  the  locus  of  the  fourth  vertex 
is  ©C(r,)  where  rj2  =2r2  -AC2.    . 

Ex.  699.  The  locus  of  the  vertex  5  of  a  a  of  given  b 
and  area  is  arc  A'BC . 

Ex.  700.  Given  b  and  (<x  +  r~P)  in  A,  the  locus  of  5 
is  arc  A5C. 


INDEX. 


PAGE 

Abbreviations vii 

Acute  angle 27,  223 

Addition 88 

Adjacent  angles 19 

Adriaan  Anthoniszoon.  .  .     130 

Metius 130 

Ahmes 130 

Alkhovarizmi 130 

Alternate  angles 33 

Altitude  of  triangle 46 

of  cone 206 

of  cylinder 208 

of  parallelogram 46 

of  pyramid 164 

of  triangle 46 

Ambiguous  case 74,  261 

Analysis 263 

Angle 16 

acute ' 27,  223 

at  center 79 

degree  of 132 

dihedral 249 

exterior 96 

exterior  of 17 

greater 73 

inscribed 55 

interior  of 17 

obtuse 28,  223 

of  two  great  circles ....    199 

polyhedral 248 

right 19 

sides  of 17 

size  of 131 

spherical 199,  218 

tanchord 60 

tetrahedral 249 


pa»b 

Angle  of  triangle 18,  219 

trihedral 249 

vertex  of 17 

Angles,  adjacent 19,  220 

alternate  exterior 33 

alternate  interior 33 

complemental 277 

corresponding 33 

exterior 33 

interior 33 

of  triangle 18 

sum  of 94 

supplemental 27 

vertical 19, 221 

Angloid 248 

Apex 164 

of  cone 205 

Apollonius 276 

Arab 130 

Arc 78 

degree  of 132 

greater 81 

length  of 129 

Arcs,  congruent 78 

Archimedes'  assumption .  6 1 

theorem 209 

Area 115 

of  circle 132 

of  cone 207 

of  cylinder 209 

of  parallelogram 122 

of  sector 132 

of  sphere 201 

of  i- 248 

of  spherical  polygon. ...  248 
279 


2  8o 


INDEX. 


,-s  PAGE 

Area  of  A  .  .       ..........  248 

of  triangle 115 

Aryabhatta 130 

Association 1,212 

Associative 89,  90 

Assumptions 2,212 

Axis  of  circle ,  .  .  .  .  196 

of  circular  cone 205 

of  circular  cylinder.  ...  209 

of  symmetry 39 

radical 278 

Base  of  cone 206 

of  cylinder 208 

of  frustum  of  cone.  ...  207 

of  parallelogram 46 

of  prismatoid 184 

of  pyramid 164 

of  spherical  segment..  .  201 

of  triangle 46 

Betweenness 5, 213, 253 

Bible 130 

Bisection-point 38,  226 

Bisection-ray 39 

Bordage 245 

Brahmagupta 72,  245 

Calculus,  sect 87 

Cavalieri 201 

Cenquad 246 

Center  of  circle 49 

of  sphere 194 

spherical 246 

straight 50 

Centimeter 129 

Central  projection 172 

Centroid 46 

Chord  of  circle 49 

of  contact 277 

of  sphere 194 

spherical 234 

Circle 49 

of  Apollonius 276 

axis  of 196 

center  of 49 

circum 52 

diameter  of 49 

division  of 274 

escribed 70 

great 194 

in- 70 

mensuration  of 120 


Circle,  radius  of 49 

small 197 

Circles,  escribed 70 

tangent 84 

two 83 

Circular  cone 205 

cylinder 208 

Circum-center 52 

of  tetrahedron 198 

-circle 52 

-radius 52 

Circumscribed  circle 52 

polygon 274 

sphere 198 

Colunar 245 

Common  section 139 

Commutative  law.  ...    88,   89 

Compasses 259 

Complement 37,  277 

Completion 109 

Concentric  cenquads 247 

Concyclic 52 

Cone 205 

apex  of 205 

area  of 205,  207 

axis  of 205 

circular 205 

element  of 205 

frustum  of 207 

lateral  area 207 

nappes  of 205 

oblique 205 

right 205 

slant  height  of 205 

truncated 207 

volume  of 205,  207 

Congruent 15 

arcs 78 

figures 29 

triangles 19 

Construction 62,  262 

Contact,  point  of 57 

Content 114,121 

Continuity 61 

Convex 11,  238 

Coplanar 3 

Corresponding 98 

Costraight 2 

Cross 160 

Cube '. 189 

Cuboid 189 


INDEX. 


281 


PAGE 

Cyclic 56 

Cylinder,  altitude  of 208 

bases  of 208 

circular 208 

elements  of 208 

lateral  area 209 

oblique 208 

right 208 

right  section  of 208 

truncated 208,  209 

volume  of 208 

Dase 130 

Data 264 

Degree  of  angle 132 

of  arc 132 

Deltoid 40 

Determination 262 

Determine 2 

Diagonal  of  polygon 10 

Diameter  of  circle 49 

of  sphere 194 

spherical 234 

Dihedral 249 

Distributive 91 

Division 91 

external 100 

internal 100 

Dodecahedron 164 

Dual 31,  158 

Duality,  principle  of.  .  31,  158 

Edges  of  angloid 248 

of  polyhedron 163 

Element  of  cone 205 

of  cylinder 208 

Elements,  the  geometric.  1 

End-points 214 

Equal 88 

polyhedrons 184 

sects 88 

Equilateral  triangle 123 

Equivalence 109,  242 

Equivalent 109 

by  completion 109 

Escribed  circle 70 

Ex-center 70 

-radius 70 

Excess,  spherical 243 

Explemental 78 

Exterior  angle 96 


PAGE 

Exterior  angles 33 

of  angle 17 

point 11 

Euclid 113,  259 

Euclid's  postulate 35 

Euler 164 

Face 248 

Faces  of  polyhedron 163 

of  tetrahedron 163 

Figure 29 

plane 29 

Foot  of  perpendicular .  ...  142 

Frustum,  area  of 207 

of  cone 207 

of  pyramid 185 

volume  of 208 

Gauss 274 

Geodesic 212 

Geometric  elements 1 

Geometry 1 

Golden  section 103 

Graphics 64 

Great  circle 194 

Greater  angle 73 

arc 81 

sect 73 

Harmonic  division 101 

range 10 1 

Heron 126 

Hexahedron 164 

Hilbert iii 

Hindoo 130 

Hippocrates  of  Chios  ....  135 

Historical  note  on  tz 130 

Hypothenuse 38,  75 

Icoshahedron 164 

In-center 70 

-radius 70 

of  tetrahedron 198 

Inclination  of  hemiplanes  155 

of  straight  to  plane  ....  151 

Inscribed  angle 55 

polygon 274 

sphere 198 

Instruments 63 

Intercept 79 

Interior  angles 33 

of  angle 17 

point 11 


282 


INDEX. 


PAGE 

Intersection  of  loci 269 

Isosceles  triangle 28 

Joachimsthal 246 

Jones 131 

Kochansky 131 

Lambert 131 

Lateral  area  of  cone 207 

edge 164 

face 164 

of  cylinder 209 

of  truncated  cylinder.  .    209 

surface 164 

Left 221 

Length  of  arc 129 

of  circle 130 

Less 89,  220 

Lexell 241 

Leyden 130 

Lindemann 131 

Loci 269,  275 

Locus 269 

Ludolph  van  Ceulen 130 

Major  arc 78 

Mean  proportional 102 

Median  of  quadrilateral.  .  40 

of  triangle 40,  228 

Meet 139 

Meter 129 

Methods 263 

Minor  arc 78 

Minus 217 

Minute 251 

Moore 253 

Motion 31 

Multiplication 89 

Nappes 205 

Oblique 74 

cylinder 208 

to  a  plane 142 

Obtuse  angle 28,  223 

Octahedron 164 

Opposite 12,  18,  212 

points 194 

side 12 


PAGB 

Opposite  summit 163 

Orthocenter 53 

Pappus 135 

Parallelogram 42 

altitude  of 46 

base  of 46 

Parallelopiped 189 

Parallel  planes 151 

straights 32 

to  a  plane 142 

Parallels 32 

Paris 129 

Partition  of  perigon 274 

transversal.  ......    115,  167 

Pascal 87 

Pasch 6,  214 

Pass 142 

Pedal 61 

Pentagon n 

Pentahedral 249 

Pentahedron 164 

Perigon 274 

partition  of 274 

Perimeter viii 

Perpendicular  planes.  ...  155 

straights 19 

to  plane 142 

Pi  («) 13° 

Plane 29 

figure 29 

tangent 195 

Planes 1 

parallel 151 

perpendicular 155 

Plus 217 

Points 1 

of  contact 57 

opposite 194 

Polar 225,  251,  276 

straightest 225 

triangle 231 

with  respect  to  a  circle .  276 

Pole 276 

of  circle 196 

Poles 196,  225 

Polygon 10 

area  of 118 

convex it 

cyclic 56 

diagonal  of 10 


INDEX. 


283 


PAGE 

Polygon,  plane 11 

regular 56 

simple 11 

spherical 216 

Polygons,  similar 105 

Polyhedral  angles 248 

Polyhedron 163 

convex 163 

edges  of 163 

faces  of 163 

summits  of 163 

volume  of 184 

Postulate,  Euclid's 35 

Principle  of  duality 158 

Prism 189 

right 189 

Prismatoid 184 

altitude  of 184 

base  of - 184 

formula 186 

section  of 185 

top  of 184 

Prismoid 185 

Problems 62,  262 

Product 89 

Projecting  plane.  .......  156 

Projection,  central 174 

of  point  on  plane 149 

sect  on  plane 149 

Proof 262 

Proportion 98 

Ptolemy 105,  130 

Pure  spherics 212 

Pyramid 164 

altitude  of 164 

apex  of 164 

base  of 164 

frustum  of 207 

lateral  edges  of 164 

lateral  faces  of 164 

lateral  surface  of 164 

truncated 207 

volume  of 185 

Pythagoras 112 

q-pole 235 

q-radius 235 

Quadrant  of  circle 93 

of  straightest 225 

Quadrilateral 11 

median  of 40 

Quotient 91 


PAGE 

K-    •    ; 52 

Radian x^2 

Radical  axis 278 

Radius  of  circle 49 

of  sphere Ig4 

spherical 234 

Ray 9>2I6 

Reckoning 272 

Rectangle 42 

Regions 7 

Regular  polygon 56 

Rhombus 42 

Richter 130 

Right 221 

angle 19 

circular  cone 205 

circular  cylinder 208 

section  of  cylinder 208 

triangle 37,75 

Rotation 94 

Ruler 63 

Scalene 87 

Schatunovsky 172 

Schur 87 

Secant 57 

Second 251 

Sect 6,  213 

-calculus 87 

-carrier 64 

greater 73 

Section 73 

common 139 

right 208 

Sector 132 

area  of 132 

Sect- train 10,  216 

Segment,  base  of 201 

of  circle 201 

of  one  base 201 

of  sphere 201 

of  straight 201 

top  of 202 

volume  of 201 

Semicircle 77 

Sense 217 

Shanks 131 

Side  of  angle 17 

of  plane 14 

of  point  on  straight.  .  .  9 

of  polygon 10 

of  straight 9 


284 


INDEX. 


PAGE 

Similar 98 

polygons 105 

Similarity 267 

Simson 264 

Size  of  angle 131 

Skew 161 

Slant  height  of  cone 205 

Small  circle 197 

Solution  of  problems ....    262 

Sphere i94 

area  of 201 

center  of 194 

circumscribed 198 

diameter  of 194 

great  circle  of 1 94 

inscribed 198 

radius  of 194 

small  circle  of 197 

tangent 195 

volume  of 201 

Spherical  angle 199 

center 246 

chord 234 

excess 243 

polygon 216 

radius 234 

segment 201 

tangent 235 

triangle 217 

Spherics 212 

Square 42 

Steradian 252 

Steregon 252 

Straight 29 

edge 63 

Straights 1 

Straightest 212 

Successive  substitutions..   263 

Sum  of  angles 94 

of  arcs 92 

of  sects 88,  237 

Summits 163 

Supplement. 27,  223 

Supplemental  sects 229 

Surface 163 

Symbols vii 

Symmetry 39,  266 

Symmetrical  points 39 

Symtra 41 

Talmud 130 

Tanchord  angle 60 


PAGE 

Tangent 57,  235 

circles 84 

Tangent,  plane 195 

spheres 195 

straight 57.195 

straightest 235 

Tetrahedral 249 

Tetrahedron 163 

circumcenter  of 198 

edges  of 163 

faces  of 163 

summits  of 163 

volume  of 167 

Theorem 3 

of  Archimedes 209 

of  Bordage 245 

of  Brahmegupta.  ...  72,  245 

of  Euler 164 

of  Heron 126 

of  Hippocrates  of  Chios.   135 

of  Joachimsthal 246 

of  Lexell 241 

of  Moore 253 

of  Pappus 135 

of  Pascal 87 

of  Ptolemy 105 

of  Pythagoras 112 

Top  of  prismatoid 184 

of  segment 202 

Trace 90 

Translation 94,  265 

Transversal 33 

partition 115,  167 

plane 167 

Trapezoid 42 

Triangle 11 

altitudes  of 46 

angle  bisectors  of 70 

angle  of 18,  219 

area  of 115 

base  of 46 

equilateral 123 

isosceles 28,  224 

medians  of 40 

polar 231 

right 37.75 

spherical 217 

Triangles,  congruent 19 

similar 98 

symmetric 222 

Trihedral , ,   249 


INDEX. 


285 


PAGE 

Trisection-points 44 

Truncated  cone 207 

cylinder 20S 

Truncated  pyramid 207 

tetrahedra 175 

Unit 89,  122,  129,  189 

circle 131 

Vega 130 

Veronese 168 

Vertex  of  angle 17 

Vertical  angles 19 

Vertices  of  polygon 10 

Volume 166 

of  cone 205 

of  cuboid 189 


PAGE 

Volume  of  cylinder 208 

of  polyhedron 184 

of  prism 189 

of  prismatoid 186 

of  pyramid 185 

of  sphere 201 

of  spherical  segment.  .   .  201 

of  tetrahedron 167 

of  truncated  cylinder.  .  209 

Within 6 

the  sphere 194 

tetrahedron 163 

Without 6 

the  sphere 1 94 

tetrahedron 163 


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Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  00 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  2  50 

*  Michie  and  Harlow's  Practical  Astronomy 8vo,  3  00 

*  White's  Elements  of  Theoretical  and  Descriptive  Astronomy nmo,  2  00 

BOTANY, 

Davenport's  Statistical  Methods,  with  Special  Reference  to  Biological  Variation. 

i6mo,  morocco,  z  25 

Thom^  and  Bennett's  Structural  and  Physiological  Botany x6mo,  2  25 

Westermaier's  Compendium  of  General  Botany.     (Schneider.) 8vo,  2  00 

CHEMISTRY. 

^driance's  Laboratory  Calculations  and  Specific  Gravity  Tables nmo,  1   25 

Allen's  Tables  for  Iron  Analysis 8vo,  3  00 

Arnold's  Compendium  of  Chemistry.     (Mandel.) Small  8vo.  3  50 

Austen's  Notes  for  Chemical  Students nmo,  1  50 

*  Austen  and  Langworthy.      The    Occurrence    of   Aluminium   in    Vegetable 

Products,  Animal  Products,  and  Natural  Waters Svo.  2  00 

Bernadou's  Smokeless  Powder. — Nitro-cellulose,  and  Theory  of  the  Cellulose 

Molecule nmo,  2  50 

Bolton's  Quantitative  Analysis 8vo,  1  50 

*  Browning's  Introduction  to  the  Rarer  Elements ,  .8vo,  x  50 

Brush  and  Penfield's  Manual  of  Determinative  Mineralogy 8vo?  4  00 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.  (Boltwood.)  ....  8vo,  3  00 

Conn's  Indicators  and  Test-papers nmo,  2  00 

Tests  and  Reagents 8vo,  3  00 

Copeland's  Manual  of  Bacteriology.     (In  preparation.) 

Craft's  Short  Course  in  Qualitative  Chemical  Analysis.  (Schaeffer.). .    .nmo,  x  SC 

Dolezalek's  Theory    of    the    Lead    Accumulator    (Storage    Battery).     (Von 

Ende). 1 2mo.  2  50 

Drechsel's  Chemical  Reactions.     (Merrill.) nmo,  1  25 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess. ) 8vo,  4  00 

Eissler's  Modern  High  Explosives. .    8vo,  4  o© 

EfEront's  Enzymes  and  their  Apphcations.     (Prescott.) . .    8vo,  3  oo 

Erdmann's  Introduction  to  Chemical  Preparations.     (Dunlap.) nmo,  x  25 

3 


Fletcher's  Practical  Instructions  in  Quantitative  Assaying  with  the  Blowpipe 

i2mo,  morocco,  i 

Fowler's  Sewage  Works  Analyses i2mo,  2 

Fresenius's  Manual  of  Qualitative  Chemical  Analysis.     (Wells.) 8vo,  5 

Manual  of  Qualitative  Chemical  Analysis.     Parti.    Descriptive.     (Wells.) 

8vo,  3 
System   of  Instruction   in    Quantitative   Chemical  Analysis.      (Cohn.) 

2  vols 8vo,  12 

Fuertes's  Water  and  Public  Health nmo,  1 

Furman's  Manual  of  Practical  Assaying 8vo,  3 

♦Getman's  Exercises  in  Physical  Chemistry i2mo,  2 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,  1 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     (Woll.) i2mo,  2 

Hammarsten's  Text-book  of  Physiological  Chemistry.     (Mandel.) 8vo,  4 

Helm's  Principles  of  Mathematical  Chemistry.     (Morgan.) i2mo,  1 

Hering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2 

Hinds's  Inorganic  Chemistry 8vo,  3 

*  Laboratory  Manual  for  Students i2mo, 

Holleman's  Text-book  of  Inorganic  Chemistry.     (Cooper.) 8vo,  2 

Text-book  of  Organic  Chemistry.     (Walker  and  Mott.) 8vo,  2 

*  Laboratory  Manual  of  Organic  Chemistry.     (Walker.) i2mo,  1 

Hopkins's  Oil-chemists'  Handbook 8vo,  3 

Jackson's  Directions  for  Laboratory  Work  in  Physiological  Chemistry. .  8vo,  1 

Keep's  Cast  Iron 8vo,  2 

Ladd's  Manual  of  Quantitative  Chemical  Analysis nmo,  1 

Landauer's  Spectrum  Analysis.     (Tingle.) 8vo,  3 

Lassar-Cohn's  Practical  Urinary  Analysis.     (Lorenz.) nmo,  1 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

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Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz.)  i2mo,  1 

Mandel's  Handbook  for  Bio-chemical  Laboratory i2mo,  1 

*  Martin's  Laboratory  Guide  to  Qualitative  Analysis  with  the  Blowpipe . .  i2mo, 
Mason's  Water-supply.  *  (Considered  Principally  from  a  Sanitary  Standpoint.) 

3d  Edition,  Rewritten 8vo,  4 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,  1 

Meyer's  Determination  of  Radicles  in  Carbon  Compounds.     (Tingle.). . i2mo,  1 

Miller's  Manual  of  Assaying i2mo,  1 

Mixter's  Elementary  Text-book  of  Chemistry i2mo,  1 

Morgan's  Outline  of  Theory  of  Solution  and  its  Results i2mo,  1 

Elements  of  Physical  Chemistry i2mo,  2 

Morse's  Calculations  used  in  Cane-sugar  Factories i6mo,  morocco,  1 

Mulliken's  General  Method  for  the  Identification  of  Pure  Organic  Compounds. 

Vol.  I Large  8vo,  5  00 

Nichols's  Water-supply.     (Considered  mainly  from  a  Chemical  and  Sanitary 

Standpoint,  1883.) 8vo,  2  50 

O'Brine's  Laboratory  Guide  in  Chemical  Analysis 8vo,  2  00 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  00 

Ost  and  Kolbeck's  Text-book  of  Chemical  Technology.     (Lorenz — Bozart.) 

(In  preparation.) 
Ostwald's  School  of  Chemistry.     Part  One.     (Ramsey.)     (In  press.) 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo,  paper,        50 

Pictet's  The  Alkaloids  and  their  Chemical  Constitution.     (Biddle.) 8vo,  5  00 

Pinner's  Introduction  to  Organic  Chemistry.     (Austen.) i2mo»  1  50 

Poole's  Calorific  Power  of  Fuels 8vo,  3  00 

Prescott  and  Winslow's  Elements  of  Water- Bacteriology,  with  Special  Refer- 
ence to  Sanitary  Water  Analysis i2mo,  1   2s 

*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  00 

4 


Richards  and  Woodman's  Air  .Water,  and  Food  from  a  Sanitary  Standpoint .  8vo, 

Richard8's  Cost  of  Living  as  Modified  by  Sanitary  Science i2mo, 

Cost  of  Food  a  Study  in  Dietaries i2mo, 

*  Richards  and  Williams's  The  Dietary  Computer 8vo, 

Ricketts  and  Russell's  Skeleton  Notes  upon  Inorganic  Chemistry.     (Part  I. — 

Non-metallic  Elements.) 8vo,  morocco, 

Ricketts  and  Miller's  Notes  on  Assaying 8vo, 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo, 

Disinfection  and  the  Preservation  of  F*ood 8vo, 

Ruddiman's  Incompatibilities  in  Prescriptions 8vo, 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish.  (In  press.) 
Sclkowski's  Physiological  and  Pathological  Chemistry.  (Orndorff.). . .  .8vo, 
Scnimpf's  Text-book  of  Volumetric  Analysis nmo, 

Essentials  of  Volumetric  Analysis nmo, 

Spencer's  Handbook  for  Chemists  or"  Beet-sugar  Houses i6mo,  morocco. 

Handbook  for  Sugar  Manufacturers  and  their  Chemists. .  i6mo,  morocco, 
Stockbridge's  Rocks  and  Soifs 8vo, 

*  Tillman's  Elementary  Lessons  in  Heat 8vo, 

*  Descriptive  General  Chemistry 8vo, 

Treadwell's  Qualitative  Analysis.     (HalL) 8vo, 

Quantitative  Analysis.     (Hall.) 8vo, 

Turneaure  and  Russell's  Public  Water-supplies 8vo, 

Van  Deventer's  Physical  Chemistry  for  Beginners.     (Boltwood.) nmo, 

*  Walke's  Lectures  on  Explosives 8vo, 

Wassermann's  Immune  Sera:  Haemolysins,  Cytotoxins,  and  Precipitins.     (Bol- 

duan.) nmo, 

Wells's  Laboratory  Guide  in  Qualitative  Chemical  Analysis 8vo, 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineering 

Students nmo, 

Whipple's  Microscopy  of  Drinking-water 8vo, 

Wiechmann's  Sugar  Analysis Small  8vo. 

Wilson's  Cyanide  Processes nmo, 

Chlorination  Process nmo. 

Wulling's  Elementary  Course  in  Inorganic    harmaceutical  and  Medical  Chem- 
istry  nmo,    a  oo 

CIVIL  ENGINEERING. 

BRIDGES  AND    ROOFS.       HYDRAULICS.      MATERIALS    OF    ENGINEERING 
RAILWAY   ENGINEERING. 

Baker's  Engineers'  Surveying  Instruments nmo,  3  00 

Bixby's  Graphical  Computing  Table Paper  19^X24^  inches.  35 

**  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal.     (Postage, 

27  cents  additional.) 8vo,  net  3  50 

Comstock's  Field  Astronomy  for  Engineers 8vo,  2  SO 

Davis's  Elevation  and  Stadia  Tables 8vo,  I  00 

Elliott's  Engineering  for  Land  Drainage nmo,  z  so 

Practical  Farm  Drainage , nmo,  1  00 

Folwell's  Sewerage.     (Designing  and  Maintenance.) 8vo,  3  00 

Freitag's  Architectural  Engineering.     2d  Edition,  Rewritten 8vo,  3  50 

French  and  Ives's  Stereotomy 8vo,  a  50 

Goodhue's  Municipal  Improvements nmo,  1  75 

Goodrich's  Economic  Disposal  of  Towns'  Refuse 8vo,  3  SO 

Gore's  Elements  of  Geodesy 8vo,  2  50 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  00 

Hering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2  50 

Howe's  Retaining  Walls  for  Earth nmo,  1   25 

Johnson's  Theory  and  Practice  of  Surveving Small  8vo,  4  00 

Statics  by  Algebraic  and  Graphic  Methods , 8vo,  a  00 

5 


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00 

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4 

00 

5 

00 

1 

50 

4 

00 

I 

00 

I 

50 

1 

50 

3 

50 

2 

50 

1 

50 

1 

50 

Kiersted's  Sewage  Disposal nmo,  i  25 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.)  nmo,  2  00 

Mahan's  Treatise  on  Civil  Engineering.     (1873O     (Wood.) 8vo,  5  00 

*  Descriptive  Geometry 8vo,  1  50 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  2  50 

Elements  of  Sanitary  Engineering 8vo,  2  00 

Merriman  and  Brooks's  Handbook  for  Surveyors i6mo,  morocco,  2  00 

Nugent's  Plane  Surveying 8vo,  3  50 

Ogden's  Sewer  Design nmo,  2  00 

Patton's  Treatise  on  Civil  Engineering 8vo  half  leather,  7  50 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  00 

Ri deal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo,  3  50 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry 8vo,  1  50 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo,  2  50 

Sondericker's  Graphic   Statics,  wuti  Applications  to   Trusses,  Beams,  and 

Arches 8vo,  2  00 

*  Trautwine's  Civil  Engineer's  Pocket-book i6mo,  morocco,  5  00 

Wait's  Engineering  and  Architectural  Jurisprudence * 8vo,  6  00 

Sheep,  6  50 
Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture.  8vo,  5  00 

Sheep,  5  50 

Law  of  Contracts 8vo,  3  00 

Warren's  Stereotomy — Problems  in  Stone-cutting 8vo,  2  50 

Webb's  Problems  in  the  U?e  and  Adjustment  of  Engineering  Instruments. 

i6mo,  morocco,  1  25 

*  Wheeler's  Elementary  Course  of  Civil  Engineering 8vo,  4  00 

Wilson's  Topographic  Surveying 8vo,  3  So 


BRIDGES  AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges . .  8vo,  2  00 

*         Thames  River  Bridge 4to,  paper,  5  00 

Burr's  Course  on  the  Stresses  in  Bridges  and  Roof  Trusses,  Arched  Ribs,  and 

Suspension  Bridges 8vo,  3  5<> 

Du  Bois's  Mechanics  of  Engineering.     VoL  II Small  4to,    10  00 

Foster's  Treatise  on  Wooden  Trestle  Bridges 4to,  5  00 

Fowler's  Coffer-dam  Process  for  Piers 8vo,  2  50 

Greene's  Roof  Trusses .". 8vo,  1  25 

Bridge  Trusses 8vo,  2  so 

Arches  in  Wood,  Iron,  and  Stone 8vo,  2  50 

Howe's  Treatise  on  Arches 8vo,  4  00 

Design  of  Simple  Roof-trusses  in  Wood  and  Steel 8vo,  2  00 

J#hnson,  Bryan,  and  Turneaure's  Theory  and  Practice  in  the  Designing  of 

Modern  Framed  Structures Small  4to,    10  00 

Merriman  and  Jacoby's  Text-book  on  Roofs  and  Bridges: 

Parti. — Stresses  in  Simple  Trusses 8vo,  2  -50 

Part  n. — Graphic  Statics 8vo,  2  50 

Part  m.— Bridge  Design.     4th  Edition,  Rewritten 8vo,  2  50 

Part  IV.— Higher  Structures 8vo,  2  50 

Morison's  Memphis  Bridge 4to,  10  00 

Waddell's  De  Pontibus,  a  Pocket-book  for  Bridge  Engineers. . .  i6mo,  morocco,  3  00 

Specifications  for  Steel  Bridges i2mo,  1  25 

Wood's  Treatise  on  the  Theory  of  the  Construction  of  Bridges  and  Roofs .  8vo,  2  00 
Wright's  Designing  of  Draw-spans: 

Part  L  —Plate-girder  Draws 8vo,  2  50 

Part  II.— Riveted-truss  and  Pin-connected  Long-span  Draws 8vo,  2  50 

Two  parts  in  one  volume **vo,  3  50 

0 


HYDRAULICS. 

Bazin's  Experiments  upon  the  Contraction  of  the  Liquid  Vein  Issuing  from  an 

Orifice.     (Trautwine.) 8vo, 

Bovey's  Treatise  on  Hydraulics 8vo, 

Church's  Mechanics  of  Engineering 8vo, 

Diagrams  of  Mean  Velocity  of  Water  in  Open  Channels paper, 

Coffin's  Graphical  Solution  of  Hydraulic  Problems i6mo,  morocco, 

Flather's  Dynamometers,  and  the  Measurement  of  Power nmo, 

FolwelTs  Water-supply  Engineering 8vo, 

Frizell's  Water-power 8vo, 

Fuertes's  Water  and  Public  Health nmo, 

Water-filtration  Works nmo, 

Ganguillet  and  Kutter's  General  Formula  for  the  Uniform  Flow  of  Water  in 

Rivers  and  Other  Channels.     (Hering  and  Trautwine.) 8vo, 

Hazen's  Filtration  of  Public  Water-supply 8vo, 

Hazlehurst's  Towers  and  Tanks  for  Water- works 8vo, 

Herschel's  115  Experiments  on  the  Carrying  Capacity  of  Large,  Riveted,  Metal 

Conduits 8vo, 

Mason's    Water-supply.     (Considered    Principally    from    a    Sanitary   Stand- 
point.)    3d  Edition,  Rewritten 8vo, 

Merriman's  Treatise  on  Hydraulics.     9th  Edition,  Rewritten 8vo, 

*  Michie's  Elements  of  Analytical  Mechanics 8vo, 

Schuyler's  Reservoirs  for  Irrigation,  Water-power,  and  Domestic   Water- 
supply Large  8vo, 

•*  Thomas  and  Watt's  Improvement  of  Riyers.     (Post.,  44  c.  additional),  4to, 

Turneaure  and  Russell's  Public  Water-supplies 8vo, 

Wegmann's  Desien  and  Construction  of  Dams 4to, 

Water-supply  of  the  City  of  New  York  from  1658  to'1895 4to, 

Weisbach's  Hydraulics  and  Hydraulic  Motors.     (Du  Bois.) 8vo, 

Wilson's  Manual  of  Irrigation  Engineering ,Small  8vo. 

Wolff's  Windmill  as  a  Prime  Mover 8vo, 

Wood's  Turbines 8vo, 

Elements  of  Analytical  Mechanics 8vo, 

MATERIALS  OF  ENGINEERING. 

Baker's  Treatise  on  Masonry  Construction 8vo, 

Roads  and  Pavements 8vo, 

Black's  United  States  Public  Works Oblong  4to, 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo, 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.     6th  Edi- 
tion, Rewritten 8vo, 

Byrne's  Highway  Construction 8vo, 

Inspection  of  the  Materials  and  Workmanship  Employed  in  Construction. 

i6mo, 

Church's  Mechanics  of  Engineering 8vo, 

Du  Bois's  Mechanics  of  Engineering.     Vol.  I Small  4to, 

Johnson's  Materials  of  Construction Large  8vo, 

Keep's  Cast  Iron 8vo, 

Lanza's  Applied  Mechanics 8vo, 

Martens's  Handbook  on  Testing  Materials.     (Henning.)     2  vols 8vo, 

Merrill's  Stones  for  Building  and  Decoration 8vo, 

Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo, 

Strength  of  Materials nmo, 

Metcalf's  Steel.     A  Manual  for  Steel-users nmo, 

Patton's  Practical  Treatise  on  Foundations 8vo, 

7 


2 

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5 

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6 

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2 

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00 

2 

00 

3 

50 

2 

50 

5 

00 

4 

00 

3 

00 

I 

25 

2 

00 

3 

00 

4 

00 

Rockwell's  Roads  and  Pavements  in  France i2mo, 

Smith's  Materials  of  Machines i2mo, 

Snow's  Principal  Species  of  Wood 8vo, 

Spalding's  Hydraulic  Cement i2mo, 

Text-book  on  Roads  and  Pavements i2mo, 

Thurston's  Materials  of  Engineering.     3  Parts 8vo, 

art  I. — Non-metallic  Materials  of  Engineering  and  Metallurgy 8vo, 

Part  II. — Iron  and  Steel 8vo, 

Part  III. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo, 

Thurston's  Text-book  of  the  Materials  of  Construction 8vo, 

Tillson's  Street  Pavements  and  Paving  Materials ; 8vo, 

Waddell's  De  Pontibus.     (A  Pocket-book  for  Bridge  Engineers.) . .  i6mo,  mor. , 

Specifications  for  Steel  Bridges i2mo, 

Wood's  Treatise  on  the  Resistance  of  Materials,  and  an  Appendix  on  the  Pres- 
ervation of  Timber 8vo, 

Elements  of  Analytical  Mechanics.. 8vo, 

Wood's  Rustless  Coatings:  Corrosion  and  Electrolysis  of  Iron  and  Steel.  .  .8vo, 

RAILWAY  ENGINEERING. 

Andrews's  Handbook  for  Street  Railway  Engineers.     3X5  inches,  morocco,    1  25 

Berg's  Buildings  and  Structures  of  American  Railroads 4to,  5  00 

Brooks's  Handbook  of  Street  Railroad  Location i6mo.  morocco,    1  50 

Butts's  Civil  Engineer's  Field-book i6mo,  morocco,    2  50 

Crandall's  Transition  Curve i6mo,  morocco,    1  50 

Railway  and  Other  Earthwork  Tables 8vo,    1  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.    i6mo,  morocco,    5  00 
Dredge's  History  of  the  Pennsylvania  Railroad:   (1879) Paper,    5  00 

*  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills,  4to,  half  mor.,    25  00 

Fisher's  Table  of  Cubic  Yards Cardboard,        25 

Godwin's  Railroad  Engineers*  Field-book  and  Explorers'  Guide i6mo,  mor.,  2  50 

Howard's  Transition  Curve  Field-book i6mo,  morocco.    1  so 

Hudson's  Tables  for  Calculating  the  Cubic  Contents  of  Excavations  and  Em- 
bankments   8vo,    1  00 

Molitor  and  Beard's  Manual  for  Resident  Engineers i6mo,  1  00 

Nagle's  Field  Manual  for  Railroad  Engineers i6mo  morocco.  3  00 

Philbrick's  Field  Manual  for  Engineers i6mo,  morocco,  3  00 

Searles  s  Field  Engineering i6mo,  morocco,  3  00 

Railroad  Spiral i6mo,  morocco,  1  50 

Taylor's  Prismoidal  Formulae  and  Earthwork 8vo,  1  50 

*  Trautwine's  Method  of  Calculating  the  Cubic  Contents  of  Excavations  and 

Embankments  by  the  Aid  of  Diagrams 8vo,  2  00 

The  Field  Practice  of  [Laying    Out    Circular    Curves    for    Railroads. 

i2mo,  morocco,  2  50 

Cross-section  Sheet Paper,  25 

Webb's  Railroad  Construction.     2d  Edition,  Rewritten i6roo.  morocco,  5  00 

Wellington's  Economic  Theory  of  the  Location  of  Railways Small  8vo,  5  00 

DRAWING. 

Barr's  Kinematics  of  Machinery 8vo,    2  50 

*  Bartlett's  Mechanical  Drawing 8vo,    3  00 

*  "  '  "         Abridged  Ed 8vo,    1  50 

Coolidge's  Manual  of  Drawing 8vo,  paper,    1  00 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  Engi- 
neers.    (In  press.) 

Durley's  Kinematics  of  Machines 8vo,    4  00 

8 


Hill's  Text-book  on  Shades  and  Shadows,  and  Perspective 8vo,    2  00 

Jamison's  Elements  of  Mechanical  Drawing.     (7n  press.) 

Jones's  Machine  Design: 

Part  I. — Kinematics  of  Machinery 8vo,    1  50 

Part  n. — Form,  Strength,  and  Proportions  of  Parts 8vo,    3  00 

MacCord's  Elements  of  Descriptive  Geometry    .        , , 8vo,    3  00 

Kinematics;   or,  Practical  Mechanism , 8vo,    5  00 

Mechanical  Drawing , . . . , 4to,    4  00 

Velocity  Diagrams 8vo,  1  50 

*  Mahan's  Descriptive  Geometry  and  Stone-cutting , 8vo,    1  50 

Industrial  Drawing.    (Thompson.) 8vo,    3  5© 

Reed's  Topographical  Drawing  and  Sketching 4to,    5  00 

Reid's  Course  in  Mechanical  Drawing 8vo,    2  00 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design . .  8vo,    3  00 

Robinson's  Principles  of  Mechanism 8vo,   3  00 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo,    2  50 

Warren's  Elements  of  Plane  and  Solid  Free-hand  Geometrical  Drawing.  .  i2mo,    1  00 

Drafting  Instruments  and  Operations i2mo,    1  25 

Manual  of  Elementary  Projection  Drawing i2mo,    1  50 

Manual  of  Elementary  Problems  in  the  Linear  Perspective  of  Form  and  5 

Shadow i2mo,    1  00 

Plane  Problems  in  Elementary  Geometry i2mo,    1  25 

Primary  Geometry i2mo,        75 

Elements  of  Descriptive  Geometry,  Shadows,  and  Perspective 8vo,    3  50 

General  Problems  of  Shades  and  Shadows 8vo,    3  00 

Elements  of  Machine  Construction  and  Drawing 8vo,    7  50 

Problems.  Theorems,  and  Examples  in  Descriptive  Geometrv 8vo,    2  50 

Weisbach's  Kinematics  and  the  Power  of  Transmission.       (Hermann  and 

Klein.)  8vo,     5  00 

Whelpley's  Practical  Instruction  in  the  Art  of  Letter  Engraving i2mo,    2  00 

Wilson's  Topographic  Surveying 8vo,    3  50 

Free-hand  Perspective 8vo,    2  50 

Free-hand  Lettering 8vo.    1  00 

Woolf's  Elementary  Course  in  Descriptive  Geometry Large  8vo,    3  00 

ELECTRICITY  AND   PHYSICS. 

Anthony  and  Brackett's  Text-book  of  Physics.     (Magie.) ...  .Small  8vo,  3  °° 

Anthony's  Lecture-notes  on  the  Theory  of  Electrical  Measurements i2mo,  1  00 

Benjamin's  History  of  Electricity 8vo.  3  00 

Voltaic  Cell. 8vo,  3  00 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.    (Boltwood.).  ,8vo,  3  0° 

Crehore  and  Squier's  Polarizing  Photo-chronograph 8vo,  3  00 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  .  i6mo,  morocco,  5  00 
Dolezalek's    Theory  of    the    Lead    Accumulator    (Storage    Battery).     (Von 

Ende.) i2mo,"72  50 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.) 8vo,  4  00 

Flather's  Dvnamo meters,  and  the  Measurement  of  Power i2mo,  3  00 

Gilbert's  De  Magnete.     (Mottelay.) 8vo,  2  50 

Hanchett's  Alternating  Currents  Explained i2mo,  1  00 

Hering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2  50 

Holman's  Precision  of  Measurements 8vo,  2  00 

Telescopic  Mirror-scale  Method,  Adjustments,  and  Tests.. ..  .Large  8vo,  75 

Landauer's  Spectrum  Analysis.    (Tingle.) 8vo,  3  00 

Le  ChateUer's  High-temperature  Measurements.  (Boudouard — .Burgess.  )i2mo,  3  00 

Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz.)  i2mo.  1  00 

•  Lyons's  Treatise  on  Electromagnetic  Phenomena.     Vols.  I.  and  II.  8vo,  each,  6  00 

*  Michie.     Elements  of  Wave  Motion  Relating  to  Sound  and  Light 8vo,  4  00 

9 


Niaudet's  Elementary  Treatise  on  Electric  Batteries.     (FishDack. ) lomo,  2  50 

•  Rosenberg's  Electrical  Engineering.    (Haldane  Gee — Kinzbrunner.). . .  .8vo,  1  50 

Ryan,  Norris,  and  Hoxie's  Electrical  Machinery.     VoL  L 8vo,  2  50 

Thurston's  Stationary  Steam-engines 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat . 8vo,  1  30 

Tory  and  Pitcher's  Manual  of  Laboratory  Physics Small  8vo,  2  00 

Ulke's  Modern  Electrolytic  Copper  Refining 8vo,  3  00 


LAW. 

*  Davis's  Elements  of  Law 8vo,  2  50 

*  Treatise  on  the  Military  Law  of  United  States 8vo,  7  00 

*  Sheep,  7  50 

Manual  for  Courts-martial i6mo,  morocco,  1  50 

Wait's  Engineering  and  Architectural  Jurisprudence 8vo,  6  00 

Sheep,'  6  50 
Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture     8vo,  5  00 

Sheep,  s  5o 

Law  of  Contracts 8vo,  3  00 

Winthrop's  Abridgment  of  Military  Law nmo,  2  50 

MANUFACTURES. 

Bernadou's  Smokeless  Powder — Nitro-cellulose  and  Theory  of  the  Cellulose 

Molecule i2mo,  2  50 

Bolland's  Iron  Founder i2mo,  2  50 

**  The  Iron  Founder,"  Supplement i2mo,  2  50 

Encyclopedia  of  Founding  and  Dictionary  of  Foundry  Terms  Used  in  the 

Practice  of  Moulding i2mo,  3  00 

Eissler's  Modern  High  Explosives 8vo,  4  00 

Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  00 

Fitzgerald's  Boston  Machinist i8mo,  1  00 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  1  00 

Hopkins's  Oil-chemists*  Handbook 8vo,  3  00 

Keep's  Cast  Iron 8vo,  2  50 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control.     (In  preparation.) 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo,  2  00 

Metcalfe's  Cost  of  Manufactures — And  the  Administration    of  Workshops, 

Public  and  Private 8vo,  5  00 

Meyer's  Modern  Locomotive  Construction 4to,  10  00 

Morse's  Calculations  used  in  Cane-sugar  Factories i6mo,  morocco,  1  50 

*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  00 

Smith's  Press-working  of  Metals 8vo,  3  00 

Spalding's  Hydraulic  Cement i2mo,  2  00 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo,  morocco,  3  00 

HandbooK  tor  sugar  Manutacturers  and  their  Chemists..  ,i6mo,  morocco,  2  00 
Thurston's  Manual  of  Steam-boilers,  their  Designs,  Construction  and  Opera- 
tion  8vo,  5  00 

*  Walke's  Lectures  on  Explosives 8vo,  4  00 

West's  American  Foundry  Practice i2mo,  2  50 

Moulder's  Text-book i2mo,  2  50 

Wiechmann's  Sugar  Analysis Small  8vo,  2  50 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3  00 

Woodbury's  Fire  Protection  of  Mills 8vo,  2  50 

Wood's  Rustless  Coatings:   Corrosion  and  Electrolysis  of  Iron  and  Steel.  .  .8vo,  4  00 

10 


MATHEMATICS. 

Baker's  Elliptic  Functions 8vo,  i  50 

*  Bass's  Elements  of  Differential  Calculus i2mo,  4  00 

Briggs's  Elements  of  Plane  Analytic  Geometry *2mov  x  °° 

Compton's  Manual  of  Logarithmic  Computations i2mo,  1  50 

Davis's  Introduction  to  the  Logic  of  Algebra 8vo,  1  50 

*  Dickson's  College  Algebra Large  i2mo,  1  50 

*  Answers  to  Dickson's  College  Algebra 8vo,  paper,  25 

*  Introduction  to  the  Theory  of  Algebraic  Equations   Large  i2mo,  1  25 

Halsted's  Elements  of  Geometry 8vo,  1  75 

Elementary  Synthetic  Geometry 8vo,  1  50 

Rational  Geometry nmo,  x  75 

•Johnson's  Three-place  Logarithmic  Tables:    Vest-pocket  size paper,  15 

100  copies  for  5  00 

*  Mounted  on  heavy  cardboard,  8  X 10  inches,  25 

10  copies  for  2  00 

Elementary  Treatise  on  the  Integral  Calculus Small  8vo,  1  50 

Curve  Tracing  in  Cartesian  Co-ordinates i2mo,  1  00 

Treatise  on  Ordinary  and  Partial  Differential  Equations Small  8vo,  3  50 

Theory -of  Errors  and  the  Method  of  Least  Squares i2mo,  1  50 

*  Theoretical  Mechanics i2mo,  3  00 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.)  i2mo,  200 

*  Ludlow  and  Bass.     Elements  of  Trigonometry  and  Logarithmic  and  Other 

Tables 8vo,  3  00 

Trigonometry  and  Tables  published  separately Each,  2  00 

*  Ludlow's  Logarithmic  and  Trigonometric  Tables 8vo,  1  00 

Maurer's  Technical  Mechanics 8vo,  4  00 

Merriman  and  Woodward's  Higher  Mathematics 8vo,  5  00 

Merriman's  Method  of  Least  Squares $vo,  2  00 

Rice  and  Johnson's  Elementary  Treatise  on  the  Differential  Calculus .  Sm.,  8vo,  3  00 

Differential  and  Integral  Calculus.     2  vols,  in  one Gmall  8vo,  2  50 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish.     (7n  press.) 

Wood's  Elements  of  Co-ordinate  Geometry 8vo,  2  00 

Trigonometry:  Analytical,  Plane,  and  Spherical nmo,  x  00 

MECHANICAL   ENGINEERING. 
MATERIALS  OF  ENGINEERING,  STEAM-ENGINES  AND  BOILERS. 

Baldwin's  Steam  Heating  for  Buildings nmo,  2  50 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

*  Bartlett's  Mechanical  Drawing 8vo,  3  00 

*  "                 "               "        Abridged  Ed 8vo.  1  50 

Benjamin's  Wrinkles  and  Recipes i2mo,  2  00 

Carpenter's  Experimental  Engineering 8vo,  6  00 

Heating  and  Ventilating  Buildings 8vo,  4  00 

Cary's  Smoke  Suppression  in  Plants  using  Bituminous  CoaL      {In  prep- 
aration.) 

Clerk's  Gas  and  Oil  Engine Small  8vo,  4  00 

Coolidge's  Manual  of  Drawing 8vo,    paper,  1  00 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  En- 
gineers.    (In  press.) 

Cromwell's  Treatise  on  Toothed  Gearing i2mo,  1  50 

Treatise  on  Belts  and  Pulleys i2mo,  1  50 

Durley's  Kinematics  of  Machines 8vo,  4  00 

Flather's  Dynamometers  and  the  Measurement  of  Power nmo,  3  00 

Rope  Driving l2mo,  2  00 

U 


Gill's  Gas  and  Fuel  Analysis  for  Engineers « . i2mo,  i 

Hall's  Car  Lubrication i2mo,  i 

Hering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2 

Hutton's  The  Gas  Engine 8vo,  5 

Jones's  Machine  Design: 

Part  I. — Kinematics  of  Machinery 8vo,  1 

Part  IL-^-Form,  Strength,  and  Proportions  of  Parts 8vo,  3 

Kent's  Mechanical  Engineer's  Pocket-book i6mo,    morocco,  5 

Kerr's  Power  and  Power  Transmission 8vo,  2 

MacCord's  Kinematics;  or,  Practical  Mechanism 8vo,  5 

Mechanical  Drawing 4to,  4 

Velocity  Diagrams 8vo,  1 

Mahan's  Industrial  Drawing.    (Thompson.) .8vo,  3 

Poole's  Calorific  Power  of  Fuels 8vo,  3 

Reid's  Course  in  Mechanical  Drawing „ 8vo.  2 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design.  .8vo,  3 

Richards's  Compressed  Air nrao,  1 

Robinson's  Principles  of  Mechanism 8vo,  3 

Smith's  Press-working  of  Metals 8vot  3 

Thurston's  Treatise  on   Friction  and    Lost  Work  in   Machinery   and   Mill 

Work * ••  8vo ,  3 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics .  nmo,  1 

Warren's  Elements  of  Machine  Construction  and  Drawing 8  <ro,  7 

Weisbach's  Kinematics  and  the  Power  of  Transmission.      Herrmann — 

Klein.) 8vo,  5 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein.).  .8vo,  5 

HydrauLcs  and  Hydraulic  Motors.     (Du  Bois.) 8vo,  5 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3 

Wood's  Turbines 8vo,  2 

MATERIALS  OF  ENGINEERING. 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.     6th  Edition, 

Reset 8vo.  7 

Church's  Mechanics  of  Engineering 8vo,  6 

Johnson'?  Materials  of  Construction Large  8vo,  6 

Keep's  Cast  Iron 8vo,  2 

Lanza's  Applied  Mechanics 8vo,  7 

Martens's  Handbook  on  Testing  Materials.     (Henning.) 8vo,  7 

Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo,  4 

Strength  of  Mater»als i2mo,  1 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo.  2 

Smith's  Materials  of  Machines nmo.  1 

Thurston's  Materials  of  Engineering 3  vols.,  Svo,  8  00 

Part   H.— Iron  and  Steel 8vo,  3  so 

Part  IH. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo  2  50 

Text-book  of  the  Materials  of  Construction 8vo,  5  00 

Wood's  Treatise  on  the  Resistance  of  Materials  and  an  Appendix  on  the 

Preservation  of  Timber 8vo,  2  00 

Elements  of  Analytical  Mechanics 8vo,  3  00 

Wood's  Rustless  Coatings:  Corrosion  and  Electrolysis  of  Iron  and  Steel..  .8vo,  4  00 

STEAM-ENGINES  AND   BOILERS. 

Carnot's  Reflections  on  the  Motive  Power  of  Heat.     (Thurston.) i2mo,  x  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  ,i6mo,  mor.,  5  co 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  1  00 

13 


Goss's  Locomotive  Sparks 8vo,  2  00 

Hemenway's  Indicator  Practice  and  Steam-engine  Economy i2mo,  2  00 

Hutton's  Mechanical  Engineering  of  Power  Plants 8vo,  5  00 

Heat  and  Heat-engines 8vo,  5  00 

Kent's  Steam-boiler  Economy 8vo,  4  00 

Kneass's  Practice  and  Theory  of  the  Injector 8vo  1  50 

MacCord's  Slide-valves 8vo,  2  00 

Meyer's  Modern  Locomotive  Construction 4to,  10  00 

Peabody's  Manual  of  the  Steam-engine  Indicator i2mo,  1  50 

Tables  of  the  Properties  of  Saturated  Steam  and  Other  Vapors 8vo,  1  00 

Thermodynamics  of  the  Steam-engine  and  Other  Heat-engines 8vo,  5  00 

Valve-gears  for  Steam-engines 8vo,  2  50 

Peabody  and  Miller's  Steam-boilers 8vo,  4  00 

Pray's  Twenty  Years  with  the  Indicator Large  8vo,  2  50 

Pupln's  Thermodynamics  of  Reversible  Cycles  in  Gases  and  Saturated  Vapors. 

(Osterberg.) nmo,  1  25 

Reagan's  Locomotives :  Simple,  Compound,  and  Electric nmo,  2  50 

Rontgen's  Principles  of  Thermodynamics.     (Du  Bois.) 8vo,  5  00 

Sinclair's  Locomotive  Engine  Running  and  Management i2mo,  2  00 

Smart's  Handbook  of  Engineering  Laboratory  Practice i2mo,  2  50 

Snow's  Steam-boiler  Practice 8vo,  3  00 

Spangler's  Valve-gears 8vo,  2  50 

Notes  on  Thermodynamics nmo,  1  00 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  00 

Thurston's  Handy  Tables 8vo,  1    50 

Manual  of  the  Steam-engine 2  vols..  8vo,  10  00 

Part  I. — History,  Structuce,  and  Theory 8vo,  6  00 

Part  H. — Design,  Construction,  and  Operation 8vo,  6  00 

Handbook  of  Engine  and  Boiler  Trials,  and  the  Use  of  the  Indicator  and 

the  Prony  Brake 8vo  5  00 

Stationary  Steam-engines 8vo,  2  50 

Steam-boiler  Explosions  in  Theory  and  in  Practice nmo  1  50 

Manual  of  Steam-boilers ,  Their  Designs,  Construction,  and  Operation .  8vo,  5  00 

Weisbach's  Heat,  Steam,  and  Steam-engines.     (Du  Bois.) 8vo,  5  00 

Whitham's  Steam-engine  Design 8vo,  5  00 

Wilson's  Treatise  on  Steam-boilers.     (Flather.) i6mo,  2  50 

Wood's  Thermodynamics  Heat  Motors,  and  Refrigerating  Machines. . .  .8vo,  4  00 


MECHANICS    AND  MACHINERY. 


Barr's  Kinematics  of  Machinery 8vo, 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo, 

Chase's  The  Art  of  Pattern-making nmo, 

Chordal. — Extracts  from  Letters nmo, 

Church's  Mechanics  of  Engineering 8vo, 

Notes  and  Examples  in  Mechanics 8vo, 

Compton's  First  Lessons  in  Metal-working nmo, 

Compton  and  De  Groodt's  The  Speed  Lathe nmo, 

Cromwell's  Treatise  on  Toothed  Gearing nmo, 

Treatise  on  Belts  and  Pulleys i2mo, 

Dana's  Text-book  of  Elementary  Mechanics  for  the   Use  of  Colleges  and 

Schools nmo, 

Dingey's  Machinery  Pattern  Making nmo, 

Dredge's  Record  of  the   Transportation   Exhibits  Building  of  the  World's 

Columbian  Exposition  of  1893 4to,  half  morocco,    5  00 

13 


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50 

7 

50 

2 

50 

2 

00 

6 

00 

2 

00 

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50 

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50 

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50 

1 

50 

1 

50 

2 

00 

Du  Bo  s's  Elementary  Principles  of  Mechanics: 

Vol.     I. — Kinematics '. 8vo, 

Vol     II.— Statics 8vo, 

Vol.  III.— Kinetics 8vo, 

Mechanics  of  Engineering.     Vol.   I Small  4to, 

Vol  II Small  4to, 

Durley's  Kinematics  of  Machines   8vo, 

Fitzgerald's  Boston  Machinist i6mo, 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2mo, 

Rope  Driving i2mo, 

Goss's  Locomotive  Sparks 8vo 

Hall's  Car  Lubrication i2mo, 

Holly's  Art  of  Saw  Filing i8mo, 

*  Johnson's  Theoretical  Mechanics i2mo, 

Statics  by  Graphic  and  Algebraic  Methods 8vo, 

Jones's  Machine  Design: 

Part  I. — Kinematics  of  Machinery 8vo, 

Part  H. — Form,  Strength,  and  Proportions  of  Parts '. 8vo, 

Kerr's  Power  and  Power  Transmission 8vo, 

Lanza's  Applied  Mechanics 8vo, 

MacCord's  Kinematics;  or,  Practical  Mechanism 8vo, 

Velocity  Diagrams  8vo, 

Maurer's  Technical  Mechanics 8vo, 

Merriman's  Text-book  on  the  Mechanics  of  Materials 8vo, 

*  Michie's  Elements  of  Analytical  Mechanics 8vo, 

Reagan's  Locomotives:  Simple,  Compound,  and  Electric i2mo, 

Reid's  Course  in  Mechanical  Drawing. 8vo, 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design.  .8vo, 

Richards's  Compressed  Air i2mo, 

Robinson's  Principles  of  Mechanism 8vo, 

Ryan,  Norris,  and  Hoxie's  Electrical  Machinery.     Vol.  1 8vo, 

Sinclair's  Locomotive-engine  Running  and  Management i2mo, 

Smith's  Press-working  of  Metals 8vo, 

Materials  of  Machines nmo, 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo, 

Thurston's  Treatise  on  Friction  and  Lost  Work  in  Machinery  and  Mill 
Work 8vo, 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics.  i2mo, 

Warren's  Elements  of  Machine  Construction  and  Drawing 8vo, 

Weisbach's    Kinematics    and    the  Power  of    Transmission.     (Herrmann — 
Klein.) 8vo, 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein.). 8vo, 
Wood's  Elements  of  Analytical  Mechanics 8vo, 

Principles  of  Elementary  Mechanics nmo, 

Turbines 8vo, 

The  World's  Columbian  Exposition  of  1893 ,    4to, 

METALLURGY. 

Egleston's  Metallurgy. of  Silver,  Gold,  and  Mercury: 

Vol.   I.— Silver 8vo,  7  50 

Vol   H. — Gold  and  Mercury 8vo,  7  S© 

**  Iles's  Lead-smelting.     (Postage  9  cents  additional.) i2mo,  2  50 

Keep's  Cast  Iron 8vo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  1  50 

Le  Chatelier's  High-temperature  Measurements.  (Boudouard — Burgess.) .  i2mo,  3  00 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo,  2  00 

Smith's  Materials  of  Machines i2mo,  1  00 

14 


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50 

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3 

50 

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4 

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3 

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2 

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2 

00 

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50 

3 

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00 

7 

50 

5 

00 

1 

50 

4 

00 

4 

00 

4 

00 

2 

5o 

2 

00 

3 

00 

I 

50 

3 

00 

2 

SO 

2 

00 

3 

00 

I 

00 

3 

00 

3 

00 

I 

00 

7 

50 

5 

00 

5 

00 

3 

00 

I 

25 

2 

50 

I 

00 

Thurston's  Materials  of  Engineering.     In  Three  Parts 8vo, 

Part  II. — Iron  and  Steel 8vo, 

Part  III. — A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 
Constituents 8vo, 

Ulke's  Modern  Electrolytic  Copper  Refining 8vo, 

MINERALOGY. 

Barringer's  Description  of  Minerals  of  Commercial  Value.     Oblong,  morocco, 

Boyd's  Resources  of  Southwest  Virginia 8vo, 

Map  of  Southwest  Virginia Pocket-book  form. 

Brush's  Manual  of  Determinative  Mineralogy.     (Penfield.) 8vo, 

Chester's  Catalogue  of  Minerals 8vo,  paper, 

Cloth, 

Dictionary  of  the  Names  of  Minerals 8vo, 

Dana's  System  of  Mineralogy Large  8vo,  half  leather, 

First  Appendix  to  Dana's  New  "System  of  Mineralogy." Large  8vo, 

Text-book  of  Mineralogy 8vo, 

Minerals  and  How  to  Study  Them nmo, 

Catalogue  of  American  Localities  of  Minerals Large  8vo, 

Manual  of  Mineralogy  and  Petrography i2mo, 

Eakle's  Mineral  Tables 8vo, 

Egleston's  Catalogue  of  Minerals  and  Synonyms 8vo, 

Hussak's  The  Determination  of  Rock-forming  Minerals.     (Smith.)  Small  8vo, 
Merrill's  Non-metallic  Minerals:  Their  Occurrence  and  Uses 8vo, 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

8vo,  paper, 

Rosenbusch's   Microscopical  Physiography   of   the   Rock-making   Minerals. 

(Iddings.) 8vo, 

•  Tillman's  Text-book  of  Important  Minerals  and  Docks 8vo, 

Williams's  Manual  of  Lithology 8vo, 

MINING. 

Beard's  Ventilation  of  Mines nmo,  a  50 

Boyd's  Resources  of  Southwest  Virginia 8vo,  3  00 

Map  of  Southwest  Virginia Pocket-book  form,  2  00 

•  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills. 

4to,  half  morocco,    25  00 

Eissler's  Modern  High  Explosives 8vo,  4  00 

Fowler's  Sewage  Works  Analyses nmo,  2  00 

Goodyear 's  Coal-mines  of  the  Western  Coast  of  the  United  States i2mo,  2  50 

Ihlseng's  Manual  of  Mining 8vo,  4  00 

**  Iles's  Lead-smelting.     (Postage  9c.  additional) i2mo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  1  50 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores 8vo,  2  00 

*  Walke's  Lectures  on  Explosives 8vo,  4  00 

Wilson's  Cyanide  Processes! ismo,  1  50 

Chlorination  Process i2mo,    1  50 

Hydraulic  and  Placer  Mining 121110,    2  00 

Treatise  on  Practical  and  Theoretical  Mine  Ventilation 12 mo     z  25 

SANITARY  SCIENCE. 

Copeland's  Manual  of  Bacteriology.     (In  preparation,) 

Folwell's  Sewerage.     (Designing,  Construction  and  Maintenance.; 8vo,  3  00 

Water-supply  Engineering 8vo,  4  00 

Fuertes's  Water  and  Public  Health Z2mo,  1  50 

Water-filtration   Works lamo,  2  50 

15 


8 

00 

3 

50 

2 

50 

3 

00 

2 

50 

3 

00 

2 

00 

4 

00 

I 

00 

I 

25 

3 

5o 

12 

50 

1 

00 

4 

00 

I 

50 

I 

00 

2 

00 

I 

25 

2 

50 

2 

00 

4 

00 

0 

50 

5 

00 

2 

00 

3 

00 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo,  i  oo 

Goodrich's  Economical  Disposal  of  Town's  Refuse Demy  8vo,  3  50 

Hazen's  Filtration  of  Public  Water-supplies 8vo,  3  00 

Kiersted's  Sewage  Disposal i2mo,  1  25 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State  . 

Control.     (In  preparation.) 
Mason's    Water-supply.     (Considered   Principally   from   a    Sanitary   Stand- 
point.)    3d  Edition,  Rewritten 8vo,  4  00 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,  1  25 

Merriman's  Elements  of  Sanitary  Engineering     . . , 8vo,  2  00 

Nichols's  Water-supply.     (Considered  Mainly  from  a  Chemical  and  Sanitary 

Standpoint.)     (1883.) 8vo,  2  50 

Ogden's  Sewer  Design i2mo,  2  00 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  with  Special  Reference 

to  Sanitary  Water  Analysis.  .    nmo,  1  25 

*  Price's  Handbook  on  Sanitation i2mo,  1  50 

Richards'.  Cost  of  Food.     A  Study  in  Dietaries. : i2mo,  1  00 

Cost  of  Living  as  Modified  by  Sanitary  Science 12 mo,  1  00 

Richards  and  Woodman  s  Air,  Water,  and  Food  from  a  Sanitary  Stand- 
point  8vo,  2  00 

*  Richards  and  Williams's  The  Dietary  Computer 8vo,  1  50 

Rideal's  Sewage  and  Bacterial  Purification  of  Sewage 8vo,  3  50 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  00 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

WoodhulTs  Notes  and  Military  Hygiene i6mo,  x  50 


MISCELLANEOUS. 

Barker's  Deep-sea  Soundings 8vo,  2  00 

Emmons's  Geological  Guide-book  of  the  Rocky  Mountain  Excursion  of  the 

International  Congress  of  Geologists , Large  8vc  1  50 

Ferrel's  Popular  Treatise  on  the  Winds 8vo  4  00 

Haines's  American  Railway  Management i2mo„  2  50 

Mott's  Composition,  Digestibility,  and  Nutritive  Value  of  Food.   Mounted  chart.  1  25 

Fallacy  of  the  Present  Theory  of  Sound i6mo  1  00 

Ricketts's  History  of  Rensselaer  Polytechnic  Institute,  1824- 1894.  Small  8vo,  3  00 

Rotherham's  Emphasized  New  Testament Large  8vo,  2  00 

Steel's  Treatise  on  the  Diseases  of  the  Dog 8vo,  3  50 

Totten's  Important  Question  in  Metrology 8vo  2  50 

The  World's  Columbian  Exposition  ot  1893 4to,  1  00 

Worcester  and  Atkinson.     Small  Hospitals,  Establishment  and  Maintenance, 
and  Suggestions  for  Hospital  Architecture,  with  Plans  for  a  Small 

Hospital i2mo,  x  25 


HEBREW  AND  CHALDEE    TEXT-BOOKS. 

Green's  Grammar  of  the  Hebrew  Language 8vo,  3  00 

Elementary  Hebrew  Grammar i2mo,  1  25 

Hebrew  Chrestomathy 8vo,  2  00 

Gesenius's  Hebrew  and  Chaldee  Lexicon  to  the  Old  Testament  Scriptures. 

(Tregelles.) Small  410,  half  morocco,  5  00 

Lett*  ris's  Hebrew  Bible 8vo,  2  25 

16 


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